please do NOT use excell ... ty .. A 0.05 significance level is used for a hypot
ID: 3258095 • Letter: P
Question
please do NOT use excell ... ty .. A 0.05 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is greatergreater than 0.5. Assume that sample data consists of 45 girls in 81 births, so the sample statistic of 5/9 results in a z score that is 1 standard deviation aboveabove 0
What is the value of the test statistic? The test statistic is___________(Type an integer or a decimal.)
f. What is the P-value? The P-value is __________(Round to four decimal places as needed.)
g. What are the critical value(s)? The critical value(s) is/are_________(Round to two decimal places as needed. Use a comma to separate answers as needed.)
h. What is the area of the critical region? The area is_______(Round to two decimal places
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.
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 ]
= 0.0556
z = (p - P) /
z = 1
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 1. We use the Normal Distribution Calculator to find P(z > 1) = 0.1587
Thus, the P-value = 0.1587
z critical = 1.96
Critical region is z greater than 1.96.
Interpret results. Since the P-value (0.1587) is greater than the significance level (0.05), we have to accept the null hypothesis.
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