Question

follow the steps below to construct a 90% confidence interval for the population mean of all the numbers of hours slept by college students the night before finals. then state whether the confidence interval you construct contradicts the reports claim. (if necessary, consult a .)
(a) click on \take sample\ to see the results for your random sample.

number of students
sample mean
sample standard deviation
17
6.875
2.012
enter the values of the sample size, the point estimate of the mean, the sample standard deviation, and the critical value you need for your 90% confidence interval. (choose the correct critical value from the table of critical values provided.) when you are done, select \compute\.
sample size:
standard error:
t_{0.005}=2.921
point estimate:
t_{0.010}=2.583
sample standard deviation:
margin of error:
t_{0.025}=2.120
critical value:
90% confidence interval:
t_{0.050}=1.746
t_{0.100}=1.337
(b) based on your sample, graph the 90% confidence interval for the population mean of all the numbers of hours slept by college students the night before finals.

• enter the values for the lower and upper limits on the graph to show your confidence interval.
• for the point (♦), enter the claim 8.29 from the report.

90% confidence interval:
0.000
10.000
5.000
0.000
2.000
4.000
6.000
8.000
10.000
(c) does the 90% confidence interval you constructed contradict the claim made in the report?
choose the best answer from the choices below.
○ no, the confidence interval does not contradict the claim. the mean of 8.29 hours from the report is inside the 90% confidence interval.
○ no, the confidence interval does not contradict the claim. the mean of 8.29 hours from the report is outside the 90% confidence interval.
○ yes, the confidence interval contradicts the claim. the mean of 8.29 hours from the report is inside the 90% confidence interval.
○ yes, the confidence interval contradicts the claim. the mean of 8.29 hours from the report is outside the 90% confidence interval.

Explanation:

Step1: Identify sample values

Sample size $n=17$, sample mean $\bar{x}=6.875$, sample standard deviation $s=2.012$. Degrees of freedom $df=17-1=16$. For 90% confidence, $\alpha/2=0.05$, so critical value $t_{0.05}=1.746$.

Step2: Calculate standard error

Standard error $SE=\frac{s}{\sqrt{n}}=\frac{2.012}{\sqrt{17}}\approx\frac{2.012}{4.123}\approx0.488$

Step3: Compute margin of error

Margin of error $ME=t_{0.05}\times SE=1.746\times0.488\approx0.852$

Step4: Find confidence interval

Lower limit: $\bar{x}-ME=6.875-0.852\approx6.023$; Upper limit: $\bar{x}+ME=6.875+0.852\approx7.727$

Step5: Check report claim

Report claim is 8.29, which is outside the interval (6.023,7.727).

Answer:

(a) Sample size:17, Point estimate:6.875, Sample standard deviation:2.012, Critical value:1.746
(b) Lower limit≈6.02, Upper limit≈7.73
(c) Yes, the confidence interval contradicts the claim. The mean of 8.29 hours from the report is outside the 90% confidence interval.