Last year, 50% of MNM, Inc., employees were female. It is believed that there ha
ID: 3369287 • Letter: L
Question
Last year, 50% of MNM, Inc., employees were female. It is believed that there has been a reduction in the percentage of females in the company. This year, in a random sample of 400 employees, 180 were female. At the.05 level, determine if there has been a significant reduction in the proportion of females. The computed test statistic is The test statistic has a p-value of Has there been a significant reduction in the percentage of females in the company? Include yes or no in your answer. Note: Use the probability distribution tables to find your p-values If possible, report the exact p-value If an exact p-value is not available, report the smallest possible range for the p-value, using the following format when you report your answer: "between lower limit p-value and upper limit p-value" For example, enter into the answer box: between .01 and.02 (But report your p-values.,)Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P > 0.50
Alternative hypothesis: P < 0.50
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.
Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).
S.D = sqrt[ P * ( 1 - P ) / n ]
S.D = 0.025
z = (p - P) / S.D
z = - 2.0
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 less than -2.0.
Thus, the P-value = 0.023.
Interpret results. Since the P-value (0.023) is less than the significance level (0.05), we cannot accept the null hypothesis.
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