g TruTalk mics for Cortana ) dent/Playerh kid:443 MATH-138-425 Statstics2017FABa
ID: 3310283 • Letter: G
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g TruTalk mics for Cortana ) dent/Playerh kid:443 MATH-138-425 Statstics2017FABanu Homework: Unit 3 Review - Pick the Situation Save Score: 0 33 of 1 pt Hw Score: 3229%, 258 of 8 pts 17.1.33 Questor, Help * the launch of an online edition The magazine plans to go ahead only it is convinced that mere than 30% of cutant seadars would shauld the company da? ssbscibe. The magazine contacted a simgle random samgl Test appropriate hypotheses and state your conclusion le of 500 current subscribers, and 179 of those survoyed expeessed interest What H, p>03 Ho P 03 "-p#03 The assumptions and condions are not met, so the tast cannot proceed Determin·the ztest statistic Select the correct choice below wo·Inecessary fil itie answer box to complete your choice @A. (Round to ho decirnal places as needed) z B. The assumptions and conditions are not met, so the test cannot proceed cck to select and enter your anwers) and then cick Check Answer Clear A PRTSCHONNE SYSRO 5 7 9Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.30
Alternative hypothesis: P > 0.30
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.0205
z = (p - P) /
z = 2.83
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 2.83. We use the Normal Distribution Calculator to find P(z > 2.83)
Thus, the P-value = 0.0023
Interpret results. Since the P-value (0.0023) is less than the significance level (0.05), we cannot accept the null hypothesis.
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