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Test Statistics Identify the test statistic that should be used for testing the following given claims.

a. The mean of the differences between IQ scores of husbands and IQ scores of their wives is

equal to 0.

b. The mean age of female CIA agents is equal to the mean age of male CIA agents.

c. The proportion of left-handed men is equal to the proportion of left-handed women.

d. The variation among pulse rates of women is equal to the variation among pulse rates of men.

Short Answer

Expert verified

The test statistics are given below:

a.\(t = \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\)

b.\(t = \frac{{\left( {{{\bar x}_F} - {{\bar x}_M}} \right) - \left( {{\mu _F} - {\mu _M}} \right)}}{{\sqrt {\frac{{s_F^2}}{{{n_F}}} + \frac{{s_M^2}}{{{n_M}}}} }}\)

c.\(t = \frac{{\left( {{{\hat p}_M} - {{\hat p}_W}} \right) - \left( {{p_M} - {p_W}} \right)}}{{\sqrt {\frac{{\bar p\bar q}}{{{n_M}}} + \frac{{\bar p\bar q}}{{{n_W}}}} }}\)

d. \(F = \frac{{s_W^2}}{{s_M^2}}\)

Step by step solution

01

Given information

Four cases are provided for the hypotheses test.

02

State the method to identify test statistic

The following steps are followed for determining the test statistic for any hypotheses test.

  1. Determine the parameter(s) being tested or compared.
  2. Note the number of groups being compared and check if they are independent or not.
  3. Check the underlying conditions of the study like normality, equal variance, etc.
  4. Note the measures known in the study.
  5. The test statistic is determined based on this knowledge.
03

Identify test statistics in each case

a.

Claim: Test that the mean difference of IQ scores between husbands and wives is 0.

The parameters being comparedare themeans of two populations of husbands and wives. The IQ scores in a population can be assumed to be normally distributed.

Note that the groups have paired observations as husbands and wives are related. Thus, it can be inferred that the two-sample paired t-test is used for testing the claim.

The related test statistic is given below:

\(t = \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\)

Here,\(\bar d,{\mu _d},{s_d},n\)are sample mean difference, true mean difference, standard deviation of sample differences, and the count of paired observations, respectively.

b.

Claim: Compare the mean age of female and male CIA agents

The parameters being comparedare themeans of two groups of female and male CIA agents. The population of ages can be assumed to be normally distributed.

Note that the groups are independent. Assume that only sample statistics are known, and sample standard deviations of groups are unequal. Thus, it can be inferred that the two-sample independent t-test willbe used for testing the claim.

The related test statistic is given below:

\(t = \frac{{\left( {{{\bar x}_F} - {{\bar x}_M}} \right) - \left( {{\mu _F} - {\mu _M}} \right)}}{{\sqrt {\frac{{s_F^2}}{{{n_F}}} + \frac{{s_M^2}}{{{n_M}}}} }}\)

Here,\({\bar x_F},{\mu _F},s_F^2,{n_F}\)are sample mean, true population mean, sample standard deviation, count of observations, respectively, for female CIA agents.

And\({\bar x_M},{\mu _M},s_M^2,{n_M}\)are sample mean, true population mean, sample standard deviation, count of observations, respectively, for male CIA agents.

c.

Claim: Compare the proportion of left-handed men and women

The parameters being compared are theproportion of two groups ofmen and women who are left-handed. The populationsare normally distributed.

Note that the groups are compared for equality of proportions. Thus, it can be inferred that the two sample z-test for proportionswillbe used for testing the claim.

The related test statistic is given below:

\(t = \frac{{\left( {{{\hat p}_M} - {{\hat p}_W}} \right) - \left( {{p_M} - {p_W}} \right)}}{{\sqrt {\frac{{\bar p\bar q}}{{{n_M}}} + \frac{{\bar p\bar q}}{{{n_W}}}} }}\)

Here,\({\hat p_W},{p_W},{n_W}\)are sample proportions, true population proportions, count of subjects in groups, respectively, for left-handed women.

And\({\hat p_M},{p_M},{n_M}\)are sample proportions, true population proportions, count of subjects in groups, respectively, for left-handed men.

Also,\(\bar p\)is the pooled proportion.

d.

Claim: Compare the variances between pulse rates of women and men

The parameters being comparedare thevariances of two groups ofpulse rates men and women. The populations of pulse rates are normally distributed.

Note that the groups are compared for equality of variances. Thus, it can be inferred that the F-test for variances willbe used for testing the claim.

The related test statistic is given below:

\(F = \frac{{s_W^2}}{{s_M^2}}\)

Here, \(s_W^2,s_M^2\)are sample variances of the pulse rates for women and men.

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Most popular questions from this chapter

Using Confidence Intervals

a. Assume that we want to use a 0.05 significance level to test the claim that p1 < p2. Which is better: A hypothesis test or a confidence interval?

b. In general, when dealing with inferences for two population proportions, which two of the following are equivalent: confidence interval method; P-value method; critical value method?

c. If we want to use a 0.05 significance level to test the claim that p1 < p2, what confidence level should we use?

d. If we test the claim in part (c) using the sample data in Exercise 1, we get this confidence interval: -0.000508 < p1 - p2 < - 0.000309. What does this confidence interval suggest about the claim?

Determining Sample Size The sample size needed to estimate the difference between two population proportions to within a margin of error E with a confidence level of 1 - a can be found by using the following expression:

\({\bf{E = }}{{\bf{z}}_{\frac{{\bf{\alpha }}}{{\bf{2}}}}}\sqrt {\frac{{{{\bf{p}}_{\bf{1}}}{{\bf{q}}_{\bf{1}}}}}{{{{\bf{n}}_{\bf{1}}}}}{\bf{ + }}\frac{{{{\bf{p}}_{\bf{2}}}{{\bf{q}}_{\bf{2}}}}}{{{{\bf{n}}_{\bf{2}}}}}} \)

Replace \({{\bf{n}}_{\bf{1}}}\;{\bf{and}}\;{{\bf{n}}_{\bf{2}}}\) by n in the preceding formula (assuming that both samples have the same size) and replace each of \({{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{q}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}\;{\bf{and}}\;{{\bf{q}}_{\bf{2}}}\)by 0.5 (because their values are not known). Solving for n results in this expression:

\({\bf{n = }}\frac{{{\bf{z}}_{\frac{{\bf{\alpha }}}{{\bf{2}}}}^{\bf{2}}}}{{{\bf{2}}{{\bf{E}}^{\bf{2}}}}}\)

Use this expression to find the size of each sample if you want to estimate the difference between the proportions of men and women who own smartphones. Assume that you want 95% confidence that your error is no more than 0.03.

Variation of Heights Use the sample data given in Exercise 3 鈥淗eights鈥 and test the claim that women and men have heights with the same variation. Use a 0.05 significance level.

Hypothesis Tests and Confidence Intervals for Hemoglobin

a. Exercise 2 includes a confidence interval. If you use the P-value method or the critical value method from Part 1 of this section to test the claim that women and men have the same mean hemoglobin levels, will the hypothesis tests and the confidence interval result in the same conclusion?

b. In general, if you conduct a hypothesis test using the methods of Part 1 of this section, will the P-value method, the critical value method, and the confidence interval method result in the same conclusion?

c. Assume that you want to use a 0.01 significance level to test the claim that the mean haemoglobin level in women is lessthan the mean hemoglobin level in men. What confidence level should be used if you want to test that claim using a confidence interval?

Degrees of Freedom

For Example 1 on page 431, we used df\( = \)smaller of\({n_1} - 1\)and\({n_2} - 1\), we got\(df = 11\), and the corresponding critical values are\(t = \pm 2.201.\)If we calculate df using Formula 9-1, we get\(df = 19.063\), and the corresponding critical values are\( \pm 2.093\). How is using the critical values of\(t = \pm 2.201\)more 鈥渃onservative鈥 than using the critical values of\( \pm 2.093\).

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