## Exercises - Hypothesis Tests for Means (One Sample)

1. In testing $H_0: \mu=20$ and $H_1: \mu<20$ at significance level .10, the sample size is 12.

1. Find the critical value, if $\sigma$ is known.

2. Find the critical value, if $\sigma$ is unknown.

1. $z=-1.282$

2. $t=-1.363$

2. Given a random sample of times for a swimming event, the $95\%$ confidence interval for the mean time for this event was found to be $168.5 \lt \mu \lt 177.4.$ Test the claim that the mean time for this event is equal to 170 seconds, using the confidence interval method with $\alpha=0.05$.

$H_0: \mu=170$,
$H_1:\mu\not=170$

Fail to reject $H_0$ because $\mu=170$ is in the confidence interval.

There is not enough evidence to reject the claim that the mean time for this event is equal to 170 seconds.

3. From a given sample of apples a $95\%$ confidence interval for the mean weight was found to be $123.7 \lt \mu \lt 148.5$ grams. Use this confidence interval to test the claim that the mean weight for this kind of apple is 150 grams.

$H_0: \mu=150$
$H_1:\mu\not=150$

Reject $H_0$ because $\mu=150$ is not in the confidence interval.

There is enough evidence to reject the claim that the mean weight for this kind of apple is 150 grams.

4. The average production of peanuts is 3000 pounds per acre. A new plant food has been developed and is tested on 46 individual plots of land. The mean yield in these test plots with the new plant food is 3120 pounds per acre. Assume the population standard deviation is 580 pounds. At $\alpha= 0.05$, test the claim that the average production has increased with the new plant food. Use the $P$-value method.

Assumptions: $n=46\geq 30$ and $\sigma$ is known.

$H_0: \mu=3000$
$H_1: \mu>3000$

Test statistic: z=1.40

$P$-value: .0808

Fail to reject the null hypothesis because the $P$-value is greater than the significance level.

There is not enough evidence to support the claim that the average production has increased with the new plant food.

5. A pasta company wants to determine the average amount of macaroni they are putting in a one-pound box. The weights (in grams) of a sample of 14 boxes is given below. $$\begin{array}{cccccccccccccc} 460&462&458&466&457&455&459&457&458&460&453&459&464&461\cr \end{array}$$ Test the claim that the mean weight of a box is equal to 454 grams (1 pound = 454 grams). Use the critical value method with $\alpha=0.01$.

Assumptions: There are no outliers and the distribution is not significantly skewed.

$H_0: \mu=454$
$H_1: \mu\not=454$

Test statistic: $t=5.722$
Critical values: $\pm 3.012$

Reject the null hypothesis because the test statistic is in the rejection region.

There is enough evidence to reject the claim that the mean weight of a box is equal to 454 grams.

6. Listed below are speeds (mi/h) of randomly selected vehicles measured on a section of highway on a weekday afternoon. Use a 0.05 significance level to test the claim that the mean speed is greater than 65 mi/h. Use the critical value method. $$\begin{array}{cccccccccccccccc} 61&64&65&65&66&66&67&67&67&68&69&69&70&73&74&88 \end{array}$$

Assumptions: There is one outlier. Remove the outlier and the remaining data set is not significantly skewed.

$H_0: \mu=65$
$H_1: \mu>65$

Test statistic: $t=2.7881$
Critical value: 1.753

Reject the null hypothesis because the test statistic is in the rejection region.

There is enough evidence to support the claim that the mean speed is greater than 65 mi/h.

7. Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period (44 weekdays), daily fees collected averaged $\$125$, with a standard deviation of$\$15$. The consultant who advised the city on this project predicted that parking revenues would average $\$130$per day. Test this prediction using the confidence interval method at significance level$0.10$. Assumptions:$n=44\geq 30H_0: \mu=130H_1: \mu\not=130t_{\alpha/2}=1.679$,$E=3.80121.2 \lt \mu \lt 128.8$Reject the null hypothesis because 130 is not in the confidence interval. There is enough evidence to reject the claim that parking revenues would average$\$130$ per day. Revenues seem to be lower.