8) In their advertisements, a new diet program would like to claim that their program results in a mean weight loss ($\mu$) of more than $10$ pounds in two weeks. State the null and alternative hypothesis.

9) Suppose we would like to estimate the mean amount of money ($\mu$) spent on books by CS students in a semester. We have the following data from $10$ randomly selected CS students: $X = \$249$ and $S = \$30$. Assume that the amount spent on books by CS students is normally distributed. To compute a 95% conﬁdence for µ, what will we use as the value of our critical point?

10) Installation of a certain hardware takes a random amount of time with a standard deviation of $5$ minutes. A computer technician installs this hardware on $64$ different computers, with the average installation time of $42$ minutes. Compute a $95\%$ conﬁdence interval for the mean installation time.

11) An exit poll of $1000$ randomly selected voters found that $515$ favored candidate A. Is the race too close to call? Answer this question by performing an appropriate test of hypothesis at $1\%$ level of signiﬁcance.

12) Bags of a certain brand of tortilla chips claim to have a net weight of $14$ ounces. The net weights actually vary slightly from bag to bag and are normally distributed with mean $\mu$. A representative of a consumer advocacy group wishes to see if there is any evidence that the mean net weight is less than advertised. For this, the representative randomly selects $16$ bags of this brand and determines the net weight of each. He ﬁnds the sample mean to be $X = 13.82$ and the sample standard deviation to be $S = 0.24$. Use these data to perform an appropriate test of hypothesis at $5\%$ signiﬁcance level.

13) The time needed for college students to complete a certain maze follows a normal distribution with a mean of $45$ seconds. To see if the mean time time $\mu$ (in seconds) is changed by vigorous exercise, we have a group of nine college students exercise vigorously for $30$ minutes and then complete the maze. The sample mean and standard deviation of the collected data is $49.2$ seconds and $3.5$ seconds respectively. Use these data to perform an appropriate test of hypothesis at $5\%$ level of signiﬁcance.

CONFIDENCE INTERVAL

chegendungu

**SOLUTION**

8)

$$H_0 : \mu = 10$$ $$vs$$ $$H_1 : \mu > 10$$

9)

We want to use a t-interval because the population standard deviation σ is not given. The critical point, in this case, is $t_{n−1,\frac{\alpha}{2}}$. So the critical value is: $$t_{9,0.025} = 2.262$$

10)

Let µ denote the mean installation time (in minutes). We want a $95\%$ conﬁdence interval for $\mu$. The following quantities are given: $1−\alpha = 0.95, n = 64, X = 42$ min, and $\sigma = 5$ min. Since $\sigma$ is known, we will use the $Z$-interval:

$$X ±Z_{\frac{\alpha}{2}}\frac{\sigma}{√n}=42±(1.96)\frac{5}{√64} = 42±1.225 = [40.8,43.2], $$

where $Z_{0.025} = 1.96$ is obtained from the normal table.Thus, the mean installation time is estimated to be between $40.8$ min and $43.2$ min, with $95\%$ conﬁdence.

11)

Let p denote the proportion of voters who favor candidate A. From the data, $\bar{p} = 515/1000 = 0.515$. We would like to test the null hypothesis $H_0 : p = 0.5$ against the two-sided alternative $H_1 : p \neq 0.5$, at $\alpha = 0.01$. Since $n = 1000$ is large, we will do a large-sample $Z$-test. The rejection rejection is $|Z| > Z_{\frac{\alpha}{2}} = 2.576$, using the normal table. The test statistic is:

$$Z =\frac{\bar{p}−p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}=\frac{0.515−0.5}{\sqrt{\frac{0.50.5}{1000}}} = 0.949$$

Since $|Z| = 0.949 < 2.576$, $H_0$ is not rejected. Thus, the race is too close to call.

12)

We need to test the null hypothesis $H_0 : \mu = 14$ against the one-sided alternative hypothesis $H_1 : \mu < 14$, at level $\alpha = 0.05$. Since $n = 16$ is small and the data are normally distributed, we will do a t-test. The rejection region is $T < −t_{n−1,\alpha} = −t_{15,0.05} = −1.753$, using the t-table. The test statistic is:

$$T =\frac{X −\mu_0}{\frac{S}{√n}} =\frac{13.82−14}{\frac{0.24}{\sqrt{16}}}= −3$$

Since $T = −3 < −1.753$, $H_0$ is rejected. Thus, there is evidence at $5\%$ signiﬁcance level that the mean weight is less than advertised.

13) We need to test the null $H_0 : \mu = 45$ against the two-sided alternative $H_1 : \mu \neq45$, at level $\alpha = 0.05$. It is given that $n = 9, X = 49.2$ and $S = 3.5$. Since $n$ is small and the data are normally distributed, we will do a t-test. The rejection region is $|T| > t_{n−1,\frac{\alpha}{2}} = t_{8,0.025} = 2.306$, using the t-table. The test statistic is:

$$T =\frac{X −\mu_0}{\frac{S}{√n}} =\frac{49.2−45}{\frac{3.5}{\sqrt{9}}}= 3.6$$

Since $|T| = 3.6 > 2.306$, $H_0$ is rejected. Thus, there is signiﬁcant evidence at $5\%$ signiﬁcance level that the mean time to complete the maze is changed after the exercise.

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