# A study, which randomly surveyed 3,500 households and drew on this information from the CRA, found that 73% of households

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A study, which randomly surveyed 3,500 households and drew on this information from the CRA (Canada Revenue Agency), found that 73% of households have conducted at least one pension rollover from an employer-sponsored retirement plan. Suppose a recent random sample of 90 households in a certain county was taken and respondents were asked whether they had ever funded a RRSP account with a rollover from an employer-sponsored retirement plan. Based on the sample data below, can you conclude at the 0.01 level of significance that the proportion of households in the county that have funded an RRSP with a rollover is different from the proportion for all households reported in the study?

72 respondents said they had funded an account; 18 respondents said they had not

Determine the null and alternative hypotheses. Choose the correct answer below.

Calculate the test statistic.
Z = ____ ?
Find the p-value.
p-value = ___ ?

Determine a conclusion.
Do not reject Ho. There is insufficient evidence to conclude that the proportion of households in the county that have funded a RRSP with a rollover is different from the proportion for all households in the study. This is a one-sample test for proportions.

Assume that 0.73 is the population proportion, which means we test the hypotheses below.

Hypothesis

H0:   p = 0.73

Ha:  p != 0.73

using the one-sample test of proportions, the z-statistic is applied.

Z  = (p^  - p)/sqrt(p * (1 - p)/n)

p^  = 72/90

p^  is the sample proportion.

p = 0.73

z = (72/90  - 0.73)/sqrt(0.73 * (1 - 0.73)/90) =  1.50

The p -value for any z-statistic can be obtained in Excel,  p(Z < z>NORM.S.DIST(1.5,TRUE)  =  0.0668.

Because the alternative hypothesis is a two-tailed test, the p-value is multiplied by 2, p = 0.0668 * 2 = 0.1336.

The p-value is quite large which means there is no evidence that the proportion is different from the hypothesized proportion.