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Chapter 10: Problem 81

Are young women delaying marriage and marrying at a later age? This question was addressed in a report issued by the Census Bureau (Associated Press, June 8 , 1991). The report stated that in 1970 (based on census results) the mean age of brides marrying for the first time was \(20.8\) years. In 1990 (based on a sample, because census results were not yet available), the mean was \(23.9\). Suppose that the 1990 sample mean had been based on a random sample of size 100 and that the sample standard deviation was \(6.4\). Is there sufficient evidence to support the claim that in 1990 women were marrying later in life than in 1970 ? Test the relevant hypotheses using \(\alpha=.01\). (Note: It is probably not reasonable to think that the distribution of age at first marriage is normal in shape.)

Short Answer

Expert verified
The final answer to the problem depends on the computed t-value and the critical t-value. If the computed t is greater than the critical t, we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that women in 1990 were marrying later than in 1970. Otherwise, we cannot reject the null hypothesis, meaning that there is not enough evidence to state that the mean age of brides marrying for the first time in 1990 was larger than in 1970.

Step by step solution

01

Set up the hypothesis

Firstly, we have to set up the null and alternative hypothesis. The null hypothesis (H0) is that the mean age of women marrying in 1990 is equal to the mean age of women marrying in 1970. The alternative hypothesis (H1) is the mean age for 1990 is greater than the mean age for 1970. Expressed in mathematical terms, H0: 碌 = 20.8 and H1: 碌 > 20.8.
02

Compute the Test Statistic

We will compute the t-value using the formula \(t = \frac{(\bar{x} - 碌_0)}{s/\sqrt{n}}\), where \( \bar{x} \) is the sample mean, \( 碌_0 \) is the hypothesized population mean, \(s \) is the sample standard deviation, and \(n \) is the sample size. Plugging in the numbers gives \(t = \frac{23.9 - 20.8}{6.4/\sqrt{100}}\).
03

Find the Critical Value and Make Decision

The critical t-value for a one-tailed test at significance level 0.01 with 99 degrees of freedom (df = n-1 = 100-1 = 99) can be found in a t-distribution table or using a statistics software. Suppose this value is t_c. If the calculated t-value is greater than the critical value, i.e., if \(t > t_c\), we reject the null hypothesis and support the claim that in 1990 women were marrying later in life than in 1970.

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Most popular questions from this chapter

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