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In Exercises 5鈥16, use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal.

Heights of Mothers and Daughters Listed below are heights (in.) of mothers and their first daughters. The data are from a journal kept by Francis Galton. (See Data Set 5 鈥淔amilyHeights鈥 in Appendix B.) Use a 0.05 significance level to test the claim that there is no difference in heights between mothers and their first daughters.

Height of Mother

68

60

61

63.5

69

64

69

64

63.5

66

Height of Daughter

68.5

60

63.5

67.5

68

65.5

69

68

64.5

63

Short Answer

Expert verified

There is insufficient evidence to reject the claim that there is no difference in heights between mothers and their first daughters.

Step by step solution

01

Given information

The heights of pairs of mothers and the first daughters are recorded.

02

Hypotheses

It is claimed thatthere is no difference in heights between mothers and their first daughters.

Corresponding to the given claim, the following hypotheses are set up:

Null Hypothesis: The mean of the difference between the heights of the mother and the first daughter is equal to 0.

\({H_0}:{\mu _d} = 0\)

Alternative Hypothesis: The mean of the difference between the heights of the mother and the first daughter is not equal to 0.

\[{H_1}:{\mu _d} \ne 0\]

The test is two-tailed.

03

Differences in the values of each matched pair

The following table shows the differences in the heights of the mother and the first daughter for each matched pair:

Mother

68

60

61

63.5

69

64

69

64

63.5

66

Daughter

68.5

60

63.5

67.5

68

65.5

69

68

64.5

63

Differences(d)

-0.5

0

-2.5

-4

1

-1.50

0

-4

-1

3

04

Mean and standard deviation of the differences

The number of pairs (n) is equal to 10.

The mean value of the differences is computed below:

\(\begin{array}{c}\bar d = \frac{{\left( { - 0.5} \right) + 0 + \ldots + 3}}{{10}}\\ = - 0.95\end{array}\)

The standard deviation of the differences is computed below:

\[\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({d_i} - \bar d)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {\left( { - 0.5} \right) - \left( { - 0.95} \right)} \right)}^2} + {{\left( {0 - \left( { - 0.95} \right)} \right)}^2} + ... + {{\left( {3 - \left( { - 0.95} \right)} \right)}^2}}}{{10 - 1}}} \\ = 2.18\end{array}\]

The mean value of the differences for the population of matched pairs \(\left( {{\mu _d}} \right)\) is considered to be equal to 0.

05

Test statistic

The value of the test statistic is computed as shown:

\(\begin{array}{c}t = \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ = \frac{{ - 0.95 - 0}}{{\frac{{2.18}}{{\sqrt {10} }}}}\\ = - 1.379\end{array}\)

The degrees of freedom are computed below:

\[\begin{array}{c}df = n - 1\\ = 10 - 1\\ = 9\end{array}\]

Referring to the t-distribution table, the critical values of t at\(\alpha = 0.05\)and degrees of freedom equal to 9 for a two-tailed test are -2.2622 and 2.2622.

The p-value of t equal to -1.379 is equal to 0.2012.

06

Conclusion

Since the test statistic value lies between the two critical values and the p-value is greater than 0.05, the null hypothesis is failed to reject.

There is insufficient evidence to reject the claim that there is no difference in heights between mothers and their first daughters.

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Most popular questions from this chapter

Testing Claims About Proportions. In Exercises 7鈥22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Denomination Effect A trial was conducted with 75 women in China given a 100-yuan bill, while another 75 women in China were given 100 yuan in the form of smaller bills (a 50-yuan bill plus two 20-yuan bills plus two 5-yuan bills). Among those given the single bill, 60 spent some or all of the money. Among those given the smaller bills, 68 spent some or all of the money (based on data from 鈥淭he Denomination Effect,鈥 by Raghubir and Srivastava, Journal of Consumer Research, Vol. 36). We want to use a 0.05 significance level to test the claim that when given a single large bill, a smaller proportion of women in China spend some or all of the money when compared to the proportion of women in China given the same amount in smaller bills.

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

c. If the significance level is changed to 0.01, does the conclusion change?

Family Heights. In Exercises 1鈥5, use the following heights (in.) The data are matched so that each column consists of heights from the same family.

Father

68.0

68.0

65.5

66.0

67.5

70.0

68.0

71.0

Mother

64.0

60.0

63.0

59.0

62.0

69.0

65.5

66.0

Son

71.0

64.0

71.0

68.0

70.0

71.0

71.7

71.0

Confidence Interval Construct a 95% confidence interval estimate of the mean height of sons. Write a brief statement that interprets the confidence interval.

We have specified a margin of error, a confidence level, and a likely range for the observed value of the sample proportion. For each exercise, obtain a sample size that will ensure a margin of error of at most the one specified (provided of course that the observed value of the sample proportion is further from 0.5than the educated guess).Obtain a sample size that will ensure a margin of error of at most the one specified.

marginoferror=0.04;confidencelevel=99%;likelyrange=0.7orless

Degrees of Freedom

For Example 1 on page 431, we used df smaller of n1-1and n2-1, we got , and the corresponding critical values aret=2.201. If we calculate df using Formula 9-1, we getdf=19.063, and the corresponding critical values are t=2.201. How is using the critical values of more 鈥渃onservative鈥 than using the critical values of 2.093.

Denomination Effect In the article 鈥淭he Denomination Effect鈥 by Priya Raghubir and Joydeep Srivastava, Journal of Consumer Research, Vol. 36, researchers reported results from studies conducted to determine whether people have different spending characteristics when they have larger bills, such as a \(20 bill, instead of smaller bills, such as twenty \)1 bills. In one trial, 89 undergraduate business students from two different colleges were randomly assigned to two different groups. In the 鈥渄ollar bill鈥 group, 46 subjects were given dollar bills; the 鈥渜uarter鈥 group consisted of 43 subjects given quarters. All subjects from both groups were given a choice of keeping the money or buying gum or mints. The article includes the claim that 鈥渕oney in a large denomination is less likely to be spent relative to an equivalent amount in smaller denominations.鈥 Test that claim using a 0.05 significance level with the following sample data from the study.

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