![\"A](\"https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2021\/07\/M07_normal_t.png\")
<\/a> Figure 7.3<\/strong>: Standard Normal Versus t Distributions. [Image Description (See Appendix D Figure 7.3)<\/a>] Click on the image to enlarge it.[\/caption]Here are the properties of a t<\/em> distribution:\r\nKey Facts: Properties of t Density Curve<\/p>\r\n\r\n<\/header>\r\n
- \r\n \t
- The total area under the curve is 1.<\/li>\r\n \t
- Bell-shaped and symmetric at 0, that is, the area to the right of any given t-score is the same as the area to the left of its negative counterpart: [latex]P(t \\: > \\: t_{\\alpha}) = P(t \\: < \\: -t_{\\alpha})[\/latex]. For example, [latex]P(t>2)=P(t<-2)[\/latex].<\/li>\r\n \t
- When the degrees of freedom <\/span>[latex]df = n-1[\/latex]\u00a0increases, the t<\/em> distribution approaches the standard normal distribution. When [latex]df = \\infty[\/latex]<\/span>, the <\/span>t<\/em> distribution becomes the standard normal.<\/span><\/li>\r\n \t
- The standard normal curve is taller and slimmer, and the t distribution\u00a0has a fatter and wider tail.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\nUnlike the standard normal table (Table II) whose main body gives left-tailed areas under the standard normal density curve, the main body of the t-<\/em>score table (Table IV) gives t<\/em>-scores, [latex]t_{\\alpha}[\/latex], which are defined in a manner analogous to [latex]z_{\\alpha}[\/latex]. That is, the t<\/em>-scores [latex]t_{\\alpha}[\/latex] is the value such that the area to its right <\/strong>is [latex]\\alpha[\/latex],\u00a0under the density curve of the t<\/em> distribution with a given degree of freedom.<\/a>\r\n\r\n[caption id=\"attachment_3071\" align=\"aligncenter\" width=\"4762\"]
![\"Part](\"https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2022\/06\/t_table_example.png\")
Figure 7.4<\/strong>: Usage of t-score Table (Table IV). [Image Description (See Appendix D Figure 7.4)<\/a>][\/caption]For example, if [latex]n=10[\/latex] and [latex]df = n-1 = 9[\/latex], then [latex]t_{0.025} = 2.262[\/latex]. That is, the t<\/em>-score 2.262 has an area of 0.025 to its right, under the t-density<\/i>\u00a0curve with 9 degrees of freedom. Notice that for each [latex]df[\/latex], the t<\/em>-table lists only 12 t<\/em>-scores. For this reason, we are often required to approximate the area to the right of a given t<\/em>-score. For example, to find the area to the right of the t<\/em>-score 1.5 under the\u00a0t<\/em> density curve with [latex]df = 9[\/latex], we first locate the t<\/em>-score 1.5, which is between 1.383 and 1.833; then, if we look at the top of the table, we see that the area to the right of 1.5 is between 0.1 and 0.05. If we use technology, for example, the R commander, we determine that the t<\/em>-score of 1.5 has a right-tailed area of 0.0839. That is, when [latex]df=9[\/latex], [latex]t_{0.0839} = 1.5[\/latex].\r\n![\"\"](\"https:\/\/openbooks.macewan.ca\/rcommander\/wp-content\/uploads\/sites\/8\/2020\/06\/activity.png\")
<\/div>\r\n\r\n Exercise: Use of the t-Score Table<\/p>\r\n\r\n<\/header>\r\n
\r\n\r\nGiven that [latex]n =15[\/latex], use the t<\/em>-score table (Table IV) to find\r\n- \r\n \t
- [latex]t_{0.025}[\/latex]<\/li>\r\n \t
- [latex]t_{0.005}[\/latex]<\/li>\r\n \t
- [latex]P(t \\geq 2.145)[\/latex], which is the area to the right of 2.145 under the t<\/em>\u00a0density curve.<\/li>\r\n \t
- [latex]P(t \\leq -2.145)[\/latex], which is the area to the left of -2.145 under the t<\/em>\u00a0density curve.<\/li>\r\n \t
- \u00a0[latex]P(t \\geq 2.5)[\/latex] , which is the area to the right of 2.5 under the t<\/em>\u00a0density curve.<\/li>\r\n<\/ol>\r\n
Show\/Hide Answer<\/summary>For [latex]n=15, df= n-1 = 14[\/latex]. Hence, we may refer to the bottom row of the table in Figure 7.4 and Figure 7.5.<\/a>\r\n\r\n[caption id=\"attachment_2829\" align=\"aligncenter\" width=\"600\"]
![\"A](\"https:\/\/openbooks.macewan.ca\/rcommander\/wp-content\/uploads\/sites\/8\/2022\/05\/tdf14_crop-1024x698.png\")
Figure 7.5<\/strong>: Critical Values of t Distribution with df=14. [Image Description (See Appendix D Figure 7.5)<\/a>][\/caption]\r\n- \r\n \t
- [latex]t_{0.025} = 2.145[\/latex]<\/li>\r\n \t
- [latex]t_{0.005} = 2.977[\/latex]<\/li>\r\n \t
- Since [latex]t_{0.025}=2.145[\/latex], it follows that [latex]P(t \\geq 2.145) = 0.025[\/latex].<\/li>\r\n \t
- First note that the t<\/em> distribution is symmetric at 0, so the area to the left of -2.145 is the same as the area to the right of 2.145. Therefore, [latex]P(t \\leq -2.145) = P(t \\geq 2.145) = 0.025[\/latex], which is the area under the t<\/em> density curve to the left of \u20132.145.<\/li>\r\n \t
- \u00a0Since 2.145 (which is [latex]t_{0.025}[\/latex]) [latex]< 2.5 < 2.624[\/latex] (which is [latex]t_{0.01}[\/latex]), the area to the right of 2.5 should be somewhere between 0.025 and 0.01. That is, [latex]0.01 < P(t \\geq 2.5) < 0.025[\/latex].<\/li>\r\n<\/ol>\r\n<\/details><\/div>\r\n<\/div>\r\n
7.2.2 One-Sample t<\/em> Interval When \u03c3<\/em> is Unknown<\/strong><\/h2>\r\nWhen the population standard deviation [latex]\\sigma[\/latex] is unknown and estimated by the sample standard deviation [latex]s[\/latex], a [latex](1-\\alpha) \\times 100\\%[\/latex] confidence interval is given by a one-sample t<\/em> interval:\r\n
\r\n\r\nAssumptions<\/strong>:\r\n- \r\n \t
- A simple random sample (SRS)<\/li>\r\n \t
- Normal population or large sample size (rule of thumb: [latex]n \\ge 30[\/latex])<\/li>\r\n \t
- The population standard deviation [latex]\\sigma[\/latex] is unknown<\/li>\r\n<\/ol>\r\nFormula<\/strong>: [latex](\\bar{x} - t_{\\alpha \/ 2} \\frac{s}{\\sqrt{n}}, \\bar{x} + t_{\\alpha \/ 2}\\frac{s}{\\sqrt{n}})[\/latex] or [latex]\\bar x \\pm t_{\\alpha\/2}\\frac{s}{\\sqrt{n}}[\/latex]\r\n\r\nInterpretation<\/strong>: We are [latex](1-\\alpha) \\times 100\\%[\/latex] confident that the interval contains the population mean [latex]\\mu[\/latex].\r\n\r\n<\/div>\r\n \r\n
\r\n Example: One-Sample t Interval<\/p>\r\n\r\n<\/header>\r\n
\r\n\r\nA computer company claims that the average lifetime of its laptops is about 4 years. A simple random sample of 36 laptops yields an average lifetime of 3.5 years with a sample standard deviation of 4.2 years.\r\n\r\nYou could use the following truncated Table IV to obtain the t-scores.<\/a>\r\n\r\n[caption id=\"attachment_3074\" align=\"aligncenter\" width=\"5070\"]![\"Part](\"https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2020\/08\/t_table_more_crop.png\")
[Image Description (See Appendix D Example 7.1)<\/a>][\/caption]\r\n- \r\n \t
- Obtain a 99% confidence interval for the population mean lifetime [latex]\\mu[\/latex].\r\nCheck the assumptions<\/strong>:\r\n
- \r\n \t
- We have a simple random sample (SRS).<\/li>\r\n \t
- We do not know whether the population is normal or not, but we have a large sample size [latex]n = 36 > 30[\/latex].<\/li>\r\n \t
- [latex]\\sigma[\/latex] is unknown and estimated by [latex]s=4.2[\/latex].<\/li>\r\n<\/ol>\r\nSteps<\/strong>:\r\n
- \r\n \t
- Find [latex]t_{\\alpha \/ 2}[\/latex]: [latex] n = 36, df = n-1 = 36-1 = 35[\/latex] [latex] 1 - \\alpha = 0.99 \\Longrightarrow \\alpha = 0.01 \\Longrightarrow \\frac{\\alpha}{2} = 0.005 \\Longrightarrow t_{\\alpha \/ 2} = t_{0.005} = 2.724[\/latex] (using Table IV).<\/li>\r\n \t
- Information: [latex]n = 36, \\bar{x} = 3.5, s = 4.2[\/latex].<\/li>\r\n \t
- Interval:\u00a0 [latex]\\begin{align*}\\bar{x} \\pm t_{\\alpha \/ 2} \\frac{s}{\\sqrt{n}}&= 3.5 \\pm 2.724 \\times \\frac{4.2}{\\sqrt{36}}=(3.5-1.9068, 3.5+1.9068 )\\\\&=(1.5932, 5.4068).\\end{align*}[\/latex]<\/li>\r\n<\/ul>\r\nInterpretation<\/strong>: We are 99% confident that the interval [latex](1.5932, 5.4068)[\/latex] contains the population mean lifetime. In other words, we are 99% confident that this computer company produces laptops with a mean lifetime somewhere between 1.5932 and 5.4068 years.<\/li>\r\n \t
- Obtain an 80% confidence interval for the population mean lifetime.\r\nSteps<\/strong>:\r\n
- \r\n \t
- Find [latex]t_{\\alpha \/ 2}[\/latex] : [latex] n = 36, df = n-1 = 36-1 = 35[\/latex] [latex] 1 - \\alpha = 0.8 \\Longrightarrow \\alpha = 0.2. \\Longrightarrow \\frac{\\alpha}{2} = 0.1 \\Longrightarrow t_{\\alpha \/ 2} = t_{0.1} = 1.306[\/latex] (using Table IV).<\/li>\r\n \t
- Information: [latex]n = 36, \\bar{x} = 3.5, s = 4.2[\/latex].<\/li>\r\n \t
- Interval:\r\n[latex]\\begin{align*}\\bar{x} \\pm t_{\\alpha \/ 2} \\frac{s}{\\sqrt{n}}&= 3.5 \\pm 1.306 \\times \\frac{4.2}{\\sqrt{36}}= ( 3.5 - 0.9142, 3.5 + 0.9142)\\\\ &= (2.5858, 4.4142 ).\\end{align*}[\/latex]<\/li>\r\n<\/ul>\r\nInterpretation<\/strong>: We are 80% confident that the interval [latex](2.5858, 4.4142)[\/latex] contains the population mean life [latex]\\mu[\/latex]. In other words, we are 80% confident that this computer company produces laptops with a mean lifetime somewhere between 2.5858 and 4.4142 years.<\/li>\r\n \t
- Does the confidence interval in part a) provide any evidence against the company\u2019s claim that the average lifetime of this brand of laptops is about 4 years?\r\nNo. Since the interval [latex](1.5932, 5.4068)[\/latex] contains 4, we can not reject the claim that the average lifetime is about 4 years.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
<\/div>\r\n\r\n Exercise: One-Sample t Interval<\/p>\r\n\r\n<\/header>\r\n
\r\n\r\nA nutrition laboratory tests 50 \u201creduced sodium\u201d hot dogs and finds the sample mean sodium content is 300 mg, with a sample standard deviation of 36 mg.\r\n- \r\n \t
- Obtain a 90% confidence interval for the mean sodium content of this brand of hot dog.<\/li>\r\n \t
- Interpret the confidence interval obtained in part (a).<\/li>\r\n \t
- Suppose that the mean sodium content of all brands of hot dogs on the market is 320 mg. Can we claim that this brand of \u201creduced sodium\u201d hot dogs has a lower average sodium content?<\/li>\r\n<\/ol>\r\n
Show\/Hide Answer<\/summary>Answers:<\/strong>\r\n
- \r\n \t
- Steps<\/strong>:\r\n
- \r\n \t
- Find [latex]t_{\\alpha \/ 2}[\/latex] : [latex]n = 50, df = n-1 = 50 -1 = 49[\/latex] [latex]1 - \\alpha = 0.9 \\Longrightarrow \\alpha = 0.1 \\Longrightarrow \\frac{\\alpha}{2} = 0.05 \\Longrightarrow t_{\\alpha \/ 2} = t_{0.05} = 1.677[\/latex] (using Table IV).<\/li>\r\n \t
- Information: [latex]n = 50, \\bar x = 300, s= 36[\/latex].<\/li>\r\n \t
- Interval:\r\n[latex]\\begin{align*}\\bar{x} \\pm t_{\\alpha \/ 2} \\frac{s}{\\sqrt{n}} &= 300 \\pm 1.677 \\times \\frac{36}{\\sqrt{50}} = (300 - 8.538, 300 + 8.538)\\\\& = (291.462, 308.538 ).\\end{align*}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
- Interpretation<\/strong>: We are 90% confident that this brand of \u201creduced sodium\u201d hot dogs has a mean sodium content somewhere between 291.462 mg and 308.538 mg.<\/li>\r\n \t
- Since the entire interval [latex](291.462, 308.538)[\/latex] is below 320 mg, we have evidence that this brand of \u201creduced sodium\u201d hot dog has a lower average sodium content than 320 mg.<\/li>\r\n<\/ol>\r\n<\/details><\/div>\r\n<\/div>","rendered":"
In practice, the population standard deviation is usually unknown. It is often estimated by the sample standard deviation<\/p>\n
[latex]s = \\sqrt{\\frac{\\sum^n_{i=1}(x_i - \\bar{x})^2}{n-1}} = \\sqrt{ \\frac{\\left( \\sum x_i ^2 \\right) - \\frac{(\\sum x_i)^2}{n} } {n-1} }.[\/latex]<\/p>\n
7.2.1\u00a0t<\/em> Distribution and t<\/em>-Score Table<\/strong><\/h2>\n
Recall the distribution of the sample mean [latex]\\bar{X}[\/latex]: if the population from which we sample is normally distributed or if the sample size is large, it follows that [latex]\\bar{X} \\sim N(\\mu, \\frac{\\sigma}{\\sqrt{n}})[\/latex]. For computational simplicity, we often transform [latex]\\bar{X}[\/latex] into the standardized variable [latex]Z = \\frac{\\bar{X} - \\mu}{\\sigma \/ \\sqrt{n}},[\/latex] which follows the standard normal distribution. However, when [latex]\\sigma[\/latex] is unknown, it is estimated with the sample standard deviation [latex]s[\/latex], and this leads to a different random variable [latex]t= \\frac{\\bar{X} - \\mu}{s \/ \\sqrt{n}}[\/latex], which follows the t distribution with a parameter called degrees of freedom [latex]df = n-1[\/latex].<\/p>\n
In general, degrees of freedom are the number of independent variables that can take arbitrary values; it equals the number of variables minus the number of relationships among the variables. For example, if two random variables, X and Y, are independent, we have [latex]df =2[\/latex]. However, if they satisfy the relationship X+Y=5, then [latex]df = 2-1=1[\/latex]. The random variable [latex]t= \\frac{\\bar{X} - \\mu}{s \/ \\sqrt{n}}[\/latex] is based on [latex]n[\/latex] random variables [latex]X_1,\u00a0 X_2, \\cdots, X_n[\/latex] with [latex]\\bar X=\\frac{X_1+X_2+\\cdots+X_n}{n}[\/latex]; therefore, we have [latex]n[\/latex] independent variables with one relationship. As a result, the degree of freedom is [latex]df=n-1[\/latex].<\/p>\n
The <\/span>t<\/em> density curve is very similar to the standard normal density curve. The following figure shows several <\/span>t<\/em> density curves with different degrees of freedom and the standard normal density curve.<\/a><\/span><\/p>\n
![\"A](\"https:\/\/openbooks.macewan.ca\/rcommander\/wp-content\/uploads\/sites\/8\/2021\/07\/M07_normal_t-300x286.png\")
<\/a>Figure 7.3<\/strong>: Standard Normal Versus t Distributions. [Image Description (See Appendix D Figure 7.3)<\/a>] Click on the image to enlarge it.<\/figcaption><\/figure>\n Here are the properties of a t<\/em> distribution:<\/p>\n
\n\n Key Facts: Properties of t Density Curve<\/p>\n<\/header>\n
\n- \n
- The total area under the curve is 1.<\/li>\n
- Bell-shaped and symmetric at 0, that is, the area to the right of any given t-score is the same as the area to the left of its negative counterpart: [latex]P(t \\: > \\: t_{\\alpha}) = P(t \\: < \\: -t_{\\alpha})[\/latex]. For example, [latex]P(t>2)=P(t<-2)[\/latex].<\/li>\n
- When the degrees of freedom <\/span>[latex]df = n-1[\/latex]\u00a0increases, the t<\/em> distribution approaches the standard normal distribution. When [latex]df = \\infty[\/latex]<\/span>, the <\/span>t<\/em> distribution becomes the standard normal.<\/span><\/li>\n
- The standard normal curve is taller and slimmer, and the t distribution\u00a0has a fatter and wider tail.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n
Unlike the standard normal table (Table II) whose main body gives left-tailed areas under the standard normal density curve, the main body of the t-<\/em>score table (Table IV) gives t<\/em>-scores, [latex]t_{\\alpha}[\/latex], which are defined in a manner analogous to [latex]z_{\\alpha}[\/latex]. That is, the t<\/em>-scores [latex]t_{\\alpha}[\/latex] is the value such that the area to its right <\/strong>is [latex]\\alpha[\/latex],\u00a0under the density curve of the t<\/em> distribution with a given degree of freedom.<\/a><\/p>\n
Figure 7.4<\/strong>: Usage of t-score Table (Table IV). [Image Description (See Appendix D Figure 7.4)<\/a>]<\/figcaption><\/figure>\n For example, if [latex]n=10[\/latex] and [latex]df = n-1 = 9[\/latex], then [latex]t_{0.025} = 2.262[\/latex]. That is, the t<\/em>-score 2.262 has an area of 0.025 to its right, under the t-density<\/i>\u00a0curve with 9 degrees of freedom. Notice that for each [latex]df[\/latex], the t<\/em>-table lists only 12 t<\/em>-scores. For this reason, we are often required to approximate the area to the right of a given t<\/em>-score. For example, to find the area to the right of the t<\/em>-score 1.5 under the\u00a0t<\/em> density curve with [latex]df = 9[\/latex], we first locate the t<\/em>-score 1.5, which is between 1.383 and 1.833; then, if we look at the top of the table, we see that the area to the right of 1.5 is between 0.1 and 0.05. If we use technology, for example, the R commander, we determine that the t<\/em>-score of 1.5 has a right-tailed area of 0.0839. That is, when [latex]df=9[\/latex], [latex]t_{0.0839} = 1.5[\/latex].<\/p>\n
<\/div>\n\n\n Exercise: Use of the t-Score Table<\/p>\n<\/header>\n
\nGiven that [latex]n =15[\/latex], use the t<\/em>-score table (Table IV) to find<\/p>\n
- \n
- [latex]t_{0.025}[\/latex]<\/li>\n
- [latex]t_{0.005}[\/latex]<\/li>\n
- [latex]P(t \\geq 2.145)[\/latex], which is the area to the right of 2.145 under the t<\/em>\u00a0density curve.<\/li>\n
- [latex]P(t \\leq -2.145)[\/latex], which is the area to the left of -2.145 under the t<\/em>\u00a0density curve.<\/li>\n
- \u00a0[latex]P(t \\geq 2.5)[\/latex] , which is the area to the right of 2.5 under the t<\/em>\u00a0density curve.<\/li>\n<\/ol>\n
\nShow\/Hide Answer<\/summary>\n
For [latex]n=15, df= n-1 = 14[\/latex]. Hence, we may refer to the bottom row of the table in Figure 7.4 and Figure 7.5.<\/a><\/p>\n
Figure 7.5<\/strong>: Critical Values of t Distribution with df=14. [Image Description (See Appendix D Figure 7.5)<\/a>]<\/figcaption><\/figure>\n - \n
- [latex]t_{0.025} = 2.145[\/latex]<\/li>\n
- [latex]t_{0.005} = 2.977[\/latex]<\/li>\n
- Since [latex]t_{0.025}=2.145[\/latex], it follows that [latex]P(t \\geq 2.145) = 0.025[\/latex].<\/li>\n
- First note that the t<\/em> distribution is symmetric at 0, so the area to the left of -2.145 is the same as the area to the right of 2.145. Therefore, [latex]P(t \\leq -2.145) = P(t \\geq 2.145) = 0.025[\/latex], which is the area under the t<\/em> density curve to the left of \u20132.145.<\/li>\n
- \u00a0Since 2.145 (which is [latex]t_{0.025}[\/latex]) [latex]< 2.5 < 2.624[\/latex] (which is [latex]t_{0.01}[\/latex]), the area to the right of 2.5 should be somewhere between 0.025 and 0.01. That is, [latex]0.01 < P(t \\geq 2.5) < 0.025[\/latex].<\/li>\n<\/ol>\n<\/details>\n<\/div>\n<\/div>\n
7.2.2 One-Sample t<\/em> Interval When \u03c3<\/em> is Unknown<\/strong><\/h2>\n
When the population standard deviation [latex]\\sigma[\/latex] is unknown and estimated by the sample standard deviation [latex]s[\/latex], a [latex](1-\\alpha) \\times 100\\%[\/latex] confidence interval is given by a one-sample t<\/em> interval:<\/p>\n
\nAssumptions<\/strong>:<\/p>\n
- \n
- A simple random sample (SRS)<\/li>\n
- Normal population or large sample size (rule of thumb: [latex]n \\ge 30[\/latex])<\/li>\n
- The population standard deviation [latex]\\sigma[\/latex] is unknown<\/li>\n<\/ol>\n
Formula<\/strong>: [latex](\\bar{x} - t_{\\alpha \/ 2} \\frac{s}{\\sqrt{n}}, \\bar{x} + t_{\\alpha \/ 2}\\frac{s}{\\sqrt{n}})[\/latex] or [latex]\\bar x \\pm t_{\\alpha\/2}\\frac{s}{\\sqrt{n}}[\/latex]<\/p>\n
Interpretation<\/strong>: We are [latex](1-\\alpha) \\times 100\\%[\/latex] confident that the interval contains the population mean [latex]\\mu[\/latex].<\/p>\n<\/div>\n
<\/p>\n
\n\n Example: One-Sample t Interval<\/p>\n<\/header>\n
\nA computer company claims that the average lifetime of its laptops is about 4 years. A simple random sample of 36 laptops yields an average lifetime of 3.5 years with a sample standard deviation of 4.2 years.<\/p>\n
You could use the following truncated Table IV to obtain the t-scores.<\/a><\/p>\n
[Image Description (See Appendix D Example 7.1)<\/a>]<\/figcaption><\/figure>\n - \n
- Obtain a 99% confidence interval for the population mean lifetime [latex]\\mu[\/latex].
\nCheck the assumptions<\/strong>:<\/p>\n- \n
- We have a simple random sample (SRS).<\/li>\n
- We do not know whether the population is normal or not, but we have a large sample size [latex]n = 36 > 30[\/latex].<\/li>\n
- [latex]\\sigma[\/latex] is unknown and estimated by [latex]s=4.2[\/latex].<\/li>\n<\/ol>\n
Steps<\/strong>:<\/p>\n
- \n
- Find [latex]t_{\\alpha \/ 2}[\/latex]: [latex]n = 36, df = n-1 = 36-1 = 35[\/latex] [latex]1 - \\alpha = 0.99 \\Longrightarrow \\alpha = 0.01 \\Longrightarrow \\frac{\\alpha}{2} = 0.005 \\Longrightarrow t_{\\alpha \/ 2} = t_{0.005} = 2.724[\/latex] (using Table IV).<\/li>\n
- Information: [latex]n = 36, \\bar{x} = 3.5, s = 4.2[\/latex].<\/li>\n
- Interval:\u00a0 [latex]\\begin{align*}\\bar{x} \\pm t_{\\alpha \/ 2} \\frac{s}{\\sqrt{n}}&= 3.5 \\pm 2.724 \\times \\frac{4.2}{\\sqrt{36}}=(3.5-1.9068, 3.5+1.9068 )\\\\&=(1.5932, 5.4068).\\end{align*}[\/latex]<\/li>\n<\/ul>\n
Interpretation<\/strong>: We are 99% confident that the interval [latex](1.5932, 5.4068)[\/latex] contains the population mean lifetime. In other words, we are 99% confident that this computer company produces laptops with a mean lifetime somewhere between 1.5932 and 5.4068 years.<\/li>\n
- Obtain an 80% confidence interval for the population mean lifetime.
\nSteps<\/strong>:<\/p>\n- \n
- Find [latex]t_{\\alpha \/ 2}[\/latex] : [latex]n = 36, df = n-1 = 36-1 = 35[\/latex] [latex]1 - \\alpha = 0.8 \\Longrightarrow \\alpha = 0.2. \\Longrightarrow \\frac{\\alpha}{2} = 0.1 \\Longrightarrow t_{\\alpha \/ 2} = t_{0.1} = 1.306[\/latex] (using Table IV).<\/li>\n
- Information: [latex]n = 36, \\bar{x} = 3.5, s = 4.2[\/latex].<\/li>\n
- Interval:
\n[latex]\\begin{align*}\\bar{x} \\pm t_{\\alpha \/ 2} \\frac{s}{\\sqrt{n}}&= 3.5 \\pm 1.306 \\times \\frac{4.2}{\\sqrt{36}}= ( 3.5 - 0.9142, 3.5 + 0.9142)\\\\ &= (2.5858, 4.4142 ).\\end{align*}[\/latex]<\/li>\n<\/ul>\nInterpretation<\/strong>: We are 80% confident that the interval [latex](2.5858, 4.4142)[\/latex] contains the population mean life [latex]\\mu[\/latex]. In other words, we are 80% confident that this computer company produces laptops with a mean lifetime somewhere between 2.5858 and 4.4142 years.<\/li>\n
- Does the confidence interval in part a) provide any evidence against the company\u2019s claim that the average lifetime of this brand of laptops is about 4 years?
\nNo. Since the interval [latex](1.5932, 5.4068)[\/latex] contains 4, we can not reject the claim that the average lifetime is about 4 years.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n\n\n Exercise: One-Sample t Interval<\/p>\n<\/header>\n
\nA nutrition laboratory tests 50 \u201creduced sodium\u201d hot dogs and finds the sample mean sodium content is 300 mg, with a sample standard deviation of 36 mg.<\/p>\n
- \n
- Obtain a 90% confidence interval for the mean sodium content of this brand of hot dog.<\/li>\n
- Interpret the confidence interval obtained in part (a).<\/li>\n
- Suppose that the mean sodium content of all brands of hot dogs on the market is 320 mg. Can we claim that this brand of \u201creduced sodium\u201d hot dogs has a lower average sodium content?<\/li>\n<\/ol>\n\n
Show\/Hide Answer<\/summary>\n
Answers:<\/strong><\/p>\n
- \n
- Steps<\/strong>:\n
- \n
- Find [latex]t_{\\alpha \/ 2}[\/latex] : [latex]n = 50, df = n-1 = 50 -1 = 49[\/latex] [latex]1 - \\alpha = 0.9 \\Longrightarrow \\alpha = 0.1 \\Longrightarrow \\frac{\\alpha}{2} = 0.05 \\Longrightarrow t_{\\alpha \/ 2} = t_{0.05} = 1.677[\/latex] (using Table IV).<\/li>\n
- Information: [latex]n = 50, \\bar x = 300, s= 36[\/latex].<\/li>\n
- Interval:
\n[latex]\\begin{align*}\\bar{x} \\pm t_{\\alpha \/ 2} \\frac{s}{\\sqrt{n}} &= 300 \\pm 1.677 \\times \\frac{36}{\\sqrt{50}} = (300 - 8.538, 300 + 8.538)\\\\& = (291.462, 308.538 ).\\end{align*}[\/latex]<\/li>\n<\/ul>\n<\/li>\n - Interpretation<\/strong>: We are 90% confident that this brand of \u201creduced sodium\u201d hot dogs has a mean sodium content somewhere between 291.462 mg and 308.538 mg.<\/li>\n
- Since the entire interval [latex](291.462, 308.538)[\/latex] is below 320 mg, we have evidence that this brand of \u201creduced sodium\u201d hot dog has a lower average sodium content than 320 mg.<\/li>\n<\/ol>\n<\/details>\n<\/div>\n<\/div>\n","protected":false},"author":19,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-792","chapter","type-chapter","status-publish","hentry"],"part":777,"_links":{"self":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/792","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/users\/19"}],"version-history":[{"count":77,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/792\/revisions"}],"predecessor-version":[{"id":4377,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/792\/revisions\/4377"}],"part":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/parts\/777"}],"metadata":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/792\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/media?parent=792"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=792"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/contributor?post=792"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/license?post=792"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Steps<\/strong>:\n
- Does the confidence interval in part a) provide any evidence against the company\u2019s claim that the average lifetime of this brand of laptops is about 4 years?
- Obtain an 80% confidence interval for the population mean lifetime.
- Obtain a 99% confidence interval for the population mean lifetime [latex]\\mu[\/latex].
- \u00a0Since 2.145 (which is [latex]t_{0.025}[\/latex]) [latex]< 2.5 < 2.624[\/latex] (which is [latex]t_{0.01}[\/latex]), the area to the right of 2.5 should be somewhere between 0.025 and 0.01. That is, [latex]0.01 < P(t \\geq 2.5) < 0.025[\/latex].<\/li>\n<\/ol>\n<\/details>\n<\/div>\n<\/div>\n
- [latex]P(t \\leq -2.145)[\/latex], which is the area to the left of -2.145 under the t<\/em>\u00a0density curve.<\/li>\n
- The standard normal curve is taller and slimmer, and the t distribution\u00a0has a fatter and wider tail.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n
- Since the entire interval [latex](291.462, 308.538)[\/latex] is below 320 mg, we have evidence that this brand of \u201creduced sodium\u201d hot dog has a lower average sodium content than 320 mg.<\/li>\r\n<\/ol>\r\n<\/details><\/div>\r\n<\/div>","rendered":"
- Steps<\/strong>:\r\n
- Does the confidence interval in part a) provide any evidence against the company\u2019s claim that the average lifetime of this brand of laptops is about 4 years?\r\nNo. Since the interval [latex](1.5932, 5.4068)[\/latex] contains 4, we can not reject the claim that the average lifetime is about 4 years.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
- Obtain an 80% confidence interval for the population mean lifetime.\r\nSteps<\/strong>:\r\n
- Obtain a 99% confidence interval for the population mean lifetime [latex]\\mu[\/latex].\r\nCheck the assumptions<\/strong>:\r\n
- \u00a0Since 2.145 (which is [latex]t_{0.025}[\/latex]) [latex]< 2.5 < 2.624[\/latex] (which is [latex]t_{0.01}[\/latex]), the area to the right of 2.5 should be somewhere between 0.025 and 0.01. That is, [latex]0.01 < P(t \\geq 2.5) < 0.025[\/latex].<\/li>\r\n<\/ol>\r\n<\/details><\/div>\r\n<\/div>\r\n
- [latex]P(t \\leq -2.145)[\/latex], which is the area to the left of -2.145 under the t<\/em>\u00a0density curve.<\/li>\r\n \t
- The standard normal curve is taller and slimmer, and the t distribution\u00a0has a fatter and wider tail.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\nUnlike the standard normal table (Table II) whose main body gives left-tailed areas under the standard normal density curve, the main body of the t-<\/em>score table (Table IV) gives t<\/em>-scores, [latex]t_{\\alpha}[\/latex], which are defined in a manner analogous to [latex]z_{\\alpha}[\/latex]. That is, the t<\/em>-scores [latex]t_{\\alpha}[\/latex] is the value such that the area to its right <\/strong>is [latex]\\alpha[\/latex],\u00a0under the density curve of the t<\/em> distribution with a given degree of freedom.<\/a>\r\n\r\n[caption id=\"attachment_3071\" align=\"aligncenter\" width=\"4762\"]