A. You are conducting a study to see if the proportion of women over 40 who regu

Question

A. You are conducting a study to see if the proportion of women over 40 who regularly have mammograms is significantly more than 0.45. You use a significance level of =0.005=0.005.

      H0:p=0.45H0:p=0.45
      H1:p>0.45H1:p>0.45

You obtain a sample of size n=326n=326 in which there are 166 successes.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

B. You wish to test the following claim (HaHa) at a significance level of =0.05=0.05.

      Ho:=53.1Ho:=53.1
      Ha:53.1Ha:53.1

You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

Column A Column B Column C Column D Column E 57.1 65.9 63.7 76 33.2 61.8 75 23.9 55.8 71.2 53.7 58.9 57.8 87.6 46.9 62.2 56.4 63.7 38.2 59.6 38.9 36.5 46.1 43.7 51.2 58.5 50.1 59.9 58.9 49.3 55.4 62.5 56.8 41.6 54.4 54.4 66.3 46.5 47.4 46.5 70 68.8 51.9 59.2 51.6 63.3 66.3 43.2 62.5 51.9 75 47.4 61 70 61.8 62.5 69.4 62.2 58.9 57.8

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.45
Alternative hypothesis: P > 0.45

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 ]

= 0.02755
z = (p - P) /

z = 2.15

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 2.15.

Thus, the P-value = 0.0158

Interpret results. Since the P-value (0.0158) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that mammograms is significantly more than 0.45.

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