CONFIDENCE INTERVAL AND HYPOTHESIS TESTING

# CONFIDENCE INTERVAL AND HYPOTHESIS TESTING

C
10 million points

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

C
10 million points

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 $90\%$ condence 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 $90\%$ condence 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 condence 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 signicant evidence at $1\%$ signicance 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 $95\%$ condence 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 $5\%$ signicance level that the scale is inaccurate.

R
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