Country Financial, a financial services company, uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time. A recent sample of 1000 adults showed 410 indicating that their financial security was more than fair. Just a year before, a sample of 900 adults showed 315 indicating that their financial security was more than fair. a. State the hypotheses that can be used to test for a significant difference between the population proportions for the two years. b. Conduct the hypothesis test and compute the p-value. At a .05 level of significance, what is your conclusion? c. What is the 95% confidence interval estimate of the difference between the two population proportions? What is your conclusion?

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alenahelenash · Accepted Answer

Find the z-test statistics.First, compute the sample proportion of recent year financial security was more than f sample proportion of just a year before financial security was more than fair and poole proportion then find the z-test statisticsFind the sample proportion of recent year financial security was more than fair.p―1=x1n1=4101000=0.41The sample proportion of recent year financial security was more than fair is 0.41.Find the sample propotion of just a year before financial security was more than fair.p―2=x2n2=315900=0.35The sample proportion of just a year before financial security was more than fair is 0.35.The pooled proportion is,p―=n1p―1+n2p―2n1+n2=(1000×0.41)+(900×0.35)1000+900=410+3151000+900=0.3816The pooled proportion is 0.3816.Find the z-test statistics.z=p―1−p―2p―(1−p―)×(1n1+1n2)=0.41−0.350.3816(1−0.3816)×(11000+1900)=2.69The z-test statistics is 2.69.Find the p-value. From cumulative standard normal distribution table, the associated probability to z-test statistics right of 2.69 is 0.9964 then subtract from 1 and multiply by 2 is,p(z&amp;lt;2.69)=2×1−[p(z&amp;lt;2.69)]=2×[1−0.9964]=2×0.0036=0.0072The p-value is 0.0072.Conclusion The p-value is less than 0.05 which is 0.0072, so the null hypothesis should be rejected at 5% significance level. There is a significant difference between the population proportions for the two years.c) Find the 95% confidence interval estimate of the difference between the two population proportions. First, compute z-critical value then find confidence interval Compute z-critical value is, From normal distribution (7) table, the z-critical value at 95% confidence level is 1.960.95%C.I.−(p―1−p―2)±zc×p―1(1−p―1)n1+p―2(1−p―2)n2=(0.41−0.35)±1.960×0.41(1−0.41)1000+0.35(1−0.35)900=0.06±0.0436=0.0164 to 0.1036The 95% cconfidence interval estimate of the difference berween the wo population proportions berween 0.0164 and 0.1036The conclusion is that there is a significant difference between the population proportions for two years due to the 95% confidence interval estimate of the difference between the two population proportions is between 0.0164 and 0.1036 does not contain 0.

Country Financial, a financial services company, uses surveys

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