I would like help with the following statistics problems. Thank you very much! w
ID: 3269077 • Letter: I
Question
I would like help with the following statistics problems. Thank you very much! wamap.org Suppose you are testing the following claim: "More than 89% of workers indicate that they are dissatisfied with their job." Express the null and alternative hypotheses in symbolic form for a hypothesis test (enter as a percentage). Ho:p decimals. Example 99%, not 0.99) (Enter percentages, not (Enter percentages, not decimals. Example 99%, not 0.99) Use the following codes to enter the following symbols: > enter>= enter>Explanation / Answer
Solution:-
1)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.89
Alternative hypothesis: P > 0.89
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.
2)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 153
Alternative hypothesis: > 153
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
3)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P = 0.36
Alternative hypothesis: P 0.36
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it 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 ]
= 0.05146
z = (p - P) /
z = 1.045
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 two-tailed test, the P-value is the probability that the z-score is less than -1.045 or greater than 1.045.
Thus, the P-value = 0.2938
Interpret results. Since the P-value (0.2938) is greater than the significance level (0.05), we cannot reject the null hypothesis.