Exam 11: Confidence Intervals and Hypothesis Tests for Means

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a. Describe the sampling distribution for the sample mean.
The sampling distribution for the sample mean can be modeled using the $t$ -distribution with 15 degrees of freedom.
b. What is the standard error?
$$S E ( \bar { y } ) = \frac { s } { \sqrt { n } } = \frac { 0.65 } { \sqrt { 16 } } = 0.1625 \mathrm { ft }$$
c. For $95 \%$ confidence, what is the margin of error?
$$t _ { n - 1 } ^ { * } \times S E ( \bar { y } ) = 2.131 \times 0.1625 f t = 0.346 f t$$
d. Based on the sample results, create the $95 \%$ confidence interval and interpret.
$$\bar { y } \pm t _ { n - 1 } ^ { * } \times S E ( \bar { y } ) = 14.8 \pm 0.346$$
The $95 \%$ confidence interval for true mean length is from $14.454$ to $15.146 \mathrm { ft }$ . We are $95 \%$ confident that the average length of metal rods from this process is between $14.454$ and $15.146 \mathrm { ft }$ .

Correct Answer:

Verified

a. Based on the sample results, find the $95 \%$ confidence interval.
The sample has a mean of $10.92$ hours and a standard deviation of $5.60$ hours.
$\bar { y } \pm t _ { n - 1 } ^ { * } \times S E ( \bar { y } ) = 10.92 \pm 2.069 \frac { 5.60 } { \sqrt { 24 } } = 10.92 \pm 2.365$
The $95 \%$ confidence interval for true mean age is $8.5$ to $13.3$ hours.
b. Interpret the results.
We are $95 \%$ confident that the average age of clients with recently paid policies is between $76.87$ and $80.33$ years.
c. Do you expect a $90 \%$ confidence interval to be wider or narrower and why?
A $90 \%$ confidence interval will be narrower because the margin of error will be smaller for a lower confidence level.
d. A previous study of a large cross-section of students in this course showed that students studies 12 hours per week. Are the results of this current study statistically different than the assumption of 12 hours per week studying? Define the hypotheses and compare the $t$ -value to the critical $t$ value, evaluating at $\alpha = 0.05$ .
Hypotheses: $\mathrm { H } _ { 0 } : \mu = 12 \mathrm { hrs } ; \mathrm { H } _ { \mathrm { A } } : \mu \neq 12 \mathrm { hrs }$
t-test: $t = \frac { 10.92 - 12 } { 5.6 / \sqrt { 24 } } = - 0.945$ ; the critical $\mathrm { t }$ value for $\mathrm { df } = 23$ at $0.05 = \pm 2.0686$
The calculated value is less than the critical value so the result is not statistically significant at $\alpha = 0.05$ . There is no evidence that the level of studying has changed.