5). Suppose a consumer advocacy group would like to conduct a survey to ﬁnd the proportion p of consumers who bought the newest generation of an MP3 player were happy with their purchase.

5a) How large a sample $n$ should they take to estimate *p* with $2\%$ margin of error and $90\%$ conﬁdence?

5b) The advocacy group took a random sample of $1000$ consumers who recently purchased this MP3 player and found that $400$ were happy with their purchase. Find a $95\%$ conﬁdence interval for *p*.

6). In order to ensure efﬁcient usage of a server, it is necessary to estimate the mean number of concurrent users. According to records, the sample mean and sample standard deviation of number of concurrent users at $100$ randomly selected times is $37.7$ and $9.2$, respectively.

6a) Construct a $90\%$ conﬁdence interval for the mean number of concurrent users.

6b) Do these data provide signiﬁcant evidence, at $1\%$ signiﬁcance level, that the mean number of concurrent users is greater than $35$?

7). To assess the accuracy of a laboratory scale, a standard weight that is known to weigh $1$ gram is repeatedly weighed $4$ times. The resulting measurements (in grams) are: $0.95, 1.02, 1.01, 0.98$. Assume that the weighing's by the scale when the true weight is $1$ gram are normally distributed with mean $\mu$.

7a) Use these data to compute a $95\%$ conﬁdence interval for $\mu$.

7b) Do these data give evidence at $5\%$ signiﬁcance level that the scale is not accurate? Answer this question by performing an appropriate test of the hypothesis.

CONFIDENCE INTERVAL

HYPOTHESIS TESTING

**SOLUTION**

5)

a) It is given that margin of error $E = 0.02$ and $\alpha = 0.10$. Using $p = 0.5$ as the conservative guess in the sample size formula gives,

$$n =\bigg[p(1−p)\bigg(\frac{Z_{\frac{\alpha}{2}}}{E}\bigg)^{2}\bigg]=\bigg[\bigg(\frac{Z_{\frac{\alpha}{2}}}{2E}\bigg)^{2}\bigg]=\bigg(\frac{1.645}{0.04}\bigg)^2= 1692.$$

b) From the data, $\bar{p} = \frac{400}{1000} = 0.40$. Since $n = 1000$ is large, the $90\%$ conﬁdence interval for $p$ is:

$$p±Z_{\frac{\alpha}{2}}\sqrt{\frac{\bar{p}(1-\bar{p})}{n}}= 0.40±1.645\sqrt{\frac{0.400.60}{1000}} = [0.375,0.425]. $$

6) Let $\mu$ denote the mean number of concurrent users in the population. It is given that $n = 100, X = 37.7$ and $S = 9.2$.

a) We want a $90\%$ conﬁdence interval for $\mu$. Since $n$ is large, we will use the large sample $Z$-interval: $$\bigg(X ±Z_{\frac{\alpha}{2}} \frac{S}{√n}\bigg)$$.

From the normal table, $Z_{0.05} = 1.645$. Thus, the desired conﬁdence interval is:

$$37.7±(1.645)\frac{9.2}{√100} = 37.7±1.5 = [36.2,39.2]$$.

b) We need to test the null $H_{0} : \mu = 35$ against the one-sided alternative $H_{1} : \mu > 35$, at level $\alpha = 0.01$. Since $n$ is large, we will do a large-sample $Z$-test. The rejection region is $Z > Z_{\alpha} = 2.33$, using the normal table. The test statistic is

$$Z =\frac{\bar{X} −\mu_{0}} {\frac{S}{√n}} =\frac{37.7-35}{\frac{9.2}{√100}}=2.93$$

Since $Z = 2.93 > 2.33$, $H_0$ is rejected. Thus, there is signiﬁcant evidence at $1\%$ signiﬁcance level that the mean number of concurrent users is greater than $35$.

7)

a) From the given data, we have: $n = 4, X = 0.99$ and $S = 0.032$. Since $n$ is small and the data are normally distributed we will use the t-interval:

$$X ±t_{n−1,\frac{\alpha}{2}}\frac{S}{√n}=0.99±3.182\frac{0.032}{2}= 0.99±0.05 = [0.94,1.04].$$

b) We need to test the null hypothesis $H_{0} : \mu = 1$ against the two-sided alternative $H_{1} : \mu \neq 1$. Since the null value of $\mu = 1$ falls in the $95\%$ conﬁdence interval computed in the previous part, it follows that the $5\%$ level test of does not reject $H_{0}$. Thus, there is no evidence at $5\%$ signiﬁcance level that the scale is inaccurate.

**Excerpt from file: **Individual School of Business MGT/434 Employment Law Affirmative Action Paper prepare a 1,050- to 1,750-word paper in which you describe the elements of affirmative action as it applies to public sector and private sector employers and how it interacts with Title VII requirements of Equal

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