7.4 Review Questions
 The value of a statistic used to estimate a parameter is called a ________ of the parameter.
 Explain the relationship between the population mean [latex]\mu[/latex], sample mean [latex]\bar X[/latex], and a value of the sample mean [latex]\bar x[/latex]. Which is the population parameter, which is a statistic, which is a point estimate, and which is an estimator?
 Suppose the average birth weight of newborn babies in Edmonton was 8 pounds in 2000. I want to know whether the average birth weight in 2010 had changed or not. Assume the population standard deviation of birth weight of newborn babies in Edmonton is [latex]\sigma = 2[/latex] pounds. Suppose a simple random sample of 100 babies in 2010 gives a mean birth weight of [latex]\bar x = 8.6[/latex] pounds.
 Obtain a 90% confidence interval for the mean birth weight of newborn babies in Edmonton in 2010.
 Interpret the confidence interval obtained in part (a).
 According to the confidence interval obtained in part (b), could you claim that the average birth weight in 2010 is different from that in 2000? If yes, how did it change?
 Determine the number of babies I should pick to guarantee that the error of [latex]\bar x[/latex] in estimating [latex]\mu[/latex] is at most 0.5 pounds with 96% confidence.
 Recalculate a 90% confidence interval using the sample size obtained in part (d) and compare it with the 90% confidence interval in part (a).
 Given that [latex]n = 15[/latex], use the tscore Table (Table IV) to find
 [latex]t_{0.025}[/latex] = ______________.
 [latex]t_{0.005}[/latex] = ______________.
 [latex]P(t\ge 2.145)[/latex] that is the area under the [latex]t[/latex] density curve to the right of 2.145.
 [latex]P(t\le 2.145)[/latex] that is the area under the t density curve to the left of 2.145.
 [latex]P(t\ge 2.5)[/latex] that is the area under the t density curve to the right of 2.5.
 A computer company claims that the average lifetime of its laptop 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.
 Obtain a 99% confidence interval for the population mean lifetime [latex]\mu[/latex].
 Obtain an 80% confidence interval for the population mean lifetime [latex]\mu[/latex].
 Does the confidence interval in part (a) support the claim that the average lifetime of this brand of laptops is about 4 years? Explain why.
 A nutrition laboratory tests 50 ”reduced sodium” hot dogs, finding that the sample mean sodium content is 300 mg, with a sample standard deviation of 36 mg.
 Obtain a 90% confidence interval for the mean sodium content of this brand of hot dog.
 Interpret the confidence interval obtained in part (a).
 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 ”reduced sodium” hot dog has a lower sodium content? Explain why.
 Suppose a 90% confidence interval for the mean weight of all newborn babies in Canada in 2015 was (5.5, 9.5) lb, which of the following statements about the interpretation of this confidence interval are correct.
 The chance that the true population mean falls in the interval (5.5, 9.5) is 0.9. (False, [latex]\mu[/latex] is fixed, and the interval is also fixed, with no randomness, no probability).
 The chance that the sample mean falls in the interval (5.5, 9.5) is 0.9. (False)
 We can be 90% confident that the interval (5.5, 9.5) contains the true population mean. (True)
 We can be 90% confident that the true population mean is somewhere between 5.5 and 9.5 lb. (True, standard way of the textbook)
 We can be 90% confident that the sample mean is somewhere between 5.5 and 9.5 lb. (False)
 Consider all samples of the same size; 90% of the sample means fall in the interval (5.5, 9.5). (False, 90% of the sample means with the same sample size are within the interval [latex]\mu\pm t_{0.05}\times \frac{s}{\sqrt{n}}[/latex].)
 Consider all samples of the same size and obtain a 90% confidence interval from each sample; 90% of those intervals contain the true population mean. (True)
Show/Hide Answer
 The value of a statistic used to estimate a parameter is called a point estimate of the parameter.
 [latex]\mu[/latex] is a population parameter that is a constant and normally unknown. [latex]\bar{X}[/latex] is a statistic and is a random variable; its value varies from sample to sample. [latex]\bar{x}[/latex] is the average of one sample; it is one value for [latex]\bar{X}[/latex] and is a point estimate of [latex]\mu[/latex].

 [latex]1  \alpha = 0.9 \Rightarrow \alpha = 1  0.9 = 0.1,z_{\alpha\text{/}2} = z_{0.05} = 1.645[/latex]
a 90% CI for mean birth weight is
[latex]\left( 8.6  1.645 \times \frac{2}{\sqrt{100}},8.6 + 1.645 \times \frac{2}{\sqrt{100}} \right) = (8.271,8.929)[/latex]
Note: the zscore [latex]z_{\alpha/2}[/latex] can be obtained at the bottom of the second page of the [latex]t[/latex]score table (Table IV).  We can be 90% confident that the mean birth weight of newborn babies in Edmonton in 2010 was between 8.271 lb and 8.929 lb.
 Yes, the entire interval is above 8lb. We can claim that the average birth weight in 2010 was greater than 8lb. The average birth weight in 2010 has increased compared to that in 2000.
 [latex]1  \alpha = 0.96,\alpha = 0.04,z_{\alpha\text{/}2} = z_{0.02} = 2.05,E = 0.5[/latex] [latex]n = \left( \frac{z_{\alpha\text{/}2} \times \sigma}{E} \right)^{2} = \left( \frac{2.05 \times 2}{0.5} \right)^{2} = 67.24[/latex]
Round up to [latex]n = 68[/latex].  [latex]1  \alpha = 0.9,\alpha = 0.1,z_{\alpha/2} = 2.05[/latex]
given n=68, a 90% CI is
[latex]\left( 8.6  1.645 \times \frac{2}{\sqrt{68}},8.6  1.645 \times \frac{2}{\sqrt{68}} \right) = (8.103,9.097)[/latex]
which is wider than the 90% interval obtained in part (a). This 90% CI is not as precise as the one in part (a) since it has a larger margin of error. Note that the sample size in part (e) is [latex]n = 68[/latex], which [latex]n = 100[/latex] in part (a), so the 90% CI in part (a) is more accurate.
 [latex]1  \alpha = 0.9 \Rightarrow \alpha = 1  0.9 = 0.1,z_{\alpha\text{/}2} = z_{0.05} = 1.645[/latex]
 Given that [latex]n = 15[/latex], use the tscore Table (Table IV) to find. [latex]df = n  1 = 15  1 = 14[/latex]
 [latex]t_{0.025} = 2.145[/latex].
 [latex]t_{0.005} = 2.977[/latex].
 [latex]P(t \geq 2.145)[/latex] that is the area under the [latex]t[/latex] density curve to the right of 2.145.
area=0.025 since [latex]t_{0.025} = 2.145[/latex], and hence [latex]P(t \geq 2.145) = 0.025[/latex].  [latex]P(t \leq  2.145)[/latex] that is the area under the t density curve to the left of 2.145.
Since [latex]t[/latex] density curve is symmetric at 0, [latex]P(t \leq  2.145) = P(t \geq 2.145) = 0.025[/latex].  [latex]P(t \geq 2.5)[/latex] that is the area under the t density curve to the right of 2.5.
Since [latex]2.415(\text{area to its right is 0.025}) < 2.5 < 2.624(\text{area to its right is 0.01})[/latex], [latex]0.01 < P(t \geq 2.5) < 0.025[/latex].

 [latex]n = 36,df = n  1 = 35,1  \alpha = 0.99,\alpha = 0.01,t_{\alpha/2} = t_{0.005} = 2.724,\bar{x} = 3.5,s = 4.2[/latex]
a 99% CI is
[latex]\left( 3.5  2.724 \times \frac{4.2}{\sqrt{36}},3.5 + 2.724 \times \frac{4.2}{\sqrt{36}} \right) = (1.593,5.407)[/latex]
Interpretation: we can be 99% confident that the average lifetime is between 1.593 and 5.407 years.  [latex]n = 36,df = n  1 = 35,1  \alpha = 0.80,\alpha = 0.2,t_{\alpha\text{/}2} = t_{0.1} = 1.306,\bar{x} = 3.5,s = 4.2[/latex]
an 80% CI is
[latex]\left( 3.5  1.306 \times \frac{4.2}{\sqrt{36}},3.5 + 1.306 \times \frac{4.2}{\sqrt{36}} \right) = (2.5858,4.4142)[/latex]
Interpretation: we can be 80% confident that the average lifetime is between 2.5858 and 4.4142 years.  Yes, since the interval contains 4. Therefore, we don’t have sufficient evidence that the population mean differs from 4.
 [latex]n = 36,df = n  1 = 35,1  \alpha = 0.99,\alpha = 0.01,t_{\alpha/2} = t_{0.005} = 2.724,\bar{x} = 3.5,s = 4.2[/latex]

 [latex]n = 50,df = n  1 = 49,1  \alpha = 0.9,\alpha = 0.1,[/latex] [latex]t_{\alpha\text{/}2} = t_{0.05} = 1.677,\bar{x} = 300,s = 36[/latex]
a 90% CI is
[latex]\left( 300  1.677 \times \frac{36}{\sqrt{50}},300 + 1.677 \times \frac{36}{\sqrt{50}} \right) = (291.462,308.538)[/latex]  Interpretation: We can be 90% confident that the population mean sodium content is somewhere between 291.462 mg to 308.538 mg
 Yes, since the entire interval is below 320 mg, which means [latex]\mu < 320[/latex]. Therefore, we can claim that this brand of “reduced sodium” hot dog has a lower sodium content.
 [latex]n = 50,df = n  1 = 49,1  \alpha = 0.9,\alpha = 0.1,[/latex] [latex]t_{\alpha\text{/}2} = t_{0.05} = 1.677,\bar{x} = 300,s = 36[/latex]
 Suppose a 90% confidence interval for the mean weight of all newborn babies in Canada in 2015 was (5.5, 9.5) lb, which of the following statements about the interpretation of this confidence interval are correct.
 False, [latex]\mu[/latex] is fixed, and the interval is also fixed, with no randomness; there is no probability concept here.
 False, similar to (a). [latex]\mu[/latex] is fixed, and the interval is also fixed; there is no randomness.
 True. If we consider all possible 90% intervals, 90% of the intervals will contain the population mean. We have 90% confidence that each of these intervals will contain the population mean.
 True, standard way of interpretation given in the textbook.
 False. The sample mean [latex]\bar{x}[/latex] is the center of the interval and hence should be in the interval for sure.
 False. Consider the distribution of the sample mean [latex]\bar{X}[/latex]. 90% of the sample means with the same sample size [latex]n[/latex] are within the interval [latex]\mu \pm t_{0.05} \times \frac{s}{\sqrt{n}}[/latex] for onesample [latex]t[/latex] interval.
 True.