# Sample Hypothesis Testing Problems

Question 1: In a classic study of infant attachment, Harlow (1959) placed infant monkeys in cages with two artificial surrogate mothers. One “mother” was made from bare wire mesh and contained a baby bottle from which the infants could feed. The other mother was made from soft terry cloth and did not provide any access to food. Harlow observed the infant monkeys and recorded how much time per day was spent with each mother. In a typical day, the infants spent a total of 18 hours clinging to one of the two mothers. If there were no preference between the two, you would expect the time to be divided evenly, with an average of µ = 9 hours for each of the mothers. However, the typical monkey spent around 15 hours per day with the terry cloth mother, indicating a strong preference for the soft, cuddly mother. Suppose a sample of n = 9 infant monkeys averaged M = 15.3 hours per day with SS = 216 with the terry cloth mother. Is this result sufficient to conclude that the monkeys spent significantly more time with the softer mother than would be expected if there were no preference? Use a two-tailed test with $$\alpha = .05$$.

Solution: We want to test following null and alternative hypotheses

\begin{align}{{H}_{0}}:\mu {=} {9}\, \\ {{H}_{A}}:\mu {\ne} {9} \\ \end{align}

Since the population standard deviation $\sigma$ is unknown, we have to use a t-test with the following formula:

$t=\frac{M-\mu }{s/\sqrt{n}}$

This corresponds to a two-tailed t-test.

$s=\sqrt{\frac{SS}{n-1}}=\sqrt{27}=5.196152$

The t-statistics is computed by the following formula:

$t=\frac{M-\mu }{s/\sqrt{n}}=\frac{15.3-9}{5.1962/\sqrt{9}}=3.6373$

The critical value for $$\alpha = 0.05$$ and for df = n- 1 = 9 -1 = 8 degrees of freedom for this two-tailed test is $$t_{c} = 2.31$$. The rejection region is given by

$R=\left\{ t:\,\,\,|t|>{2.31} \right\}$

Since $$|t| = 3.6373 {>} t_c = 2.31$$, then we reject the null hypothesis H0.

Hence, we have enough evidence to support the claim that the monkeys spent significantly more time with the softer mother than would be expected if there were no preference.

Question 2: Given a sample size of 38, with sample mean 660.3 and sample standard deviation 95.9 we are to perform the following hypothesis test.

Null Hypothesis H0: μ = 700

Alternative Hypothesis H0: μ ≠ 700

At significance level 0.05

a. Calculate the test statistics
(Tip: this is the case when we test the claim about population mean with population standard deviation not known; 95.9 is a sample standard deviation not a population standard deviation).

b. Use Table A-3 to find critical value for this test and make decision:
reject or do not reject the Null Hypothesis

Solution: a) Our interest is in testing the following null and alternative hypotheses

\begin{align}{{H}_{0}}:\mu {=} {700}\, \\ {{H}_{A}}:\mu {\ne} {700} \\ \end{align}

Since the population standard deviation $$\sigma$$ is unknown, we have to use a t-test with the following expression:

$t =\frac{\bar{X}-\mu }{s / \sqrt{n}}$

This corresponds to a two-tailed t-test. The t-statistics is computed by the following formula:

$t=\frac{\bar{X}-\mu }{s /\sqrt{n}}=\frac{{660.3}-700}{95.9/\sqrt{38}}={-2.5519}$

b) The critical value for $$\alpha = 0.05$$ and for $$df = n- 1 = 38 -1 = 37$$ degrees of freedom for this two-tailed test is $$t_{c} = 2.026$$. The rejection region is given by

$R=\left\{ t:\,\,\,|t|>{2.026} \right\}$

Since $$|t| = 2.5519 {>} t_c = 2.026$$, then we reject the null hypothesis H0.

Hence, we have enough evidence to support the claim that the population mean is different than 700.

Question 3: A real estate agent wishes to determine whether tax assessors and real estate appraisers agree on the values of homes. A random sample of the two groups appraised 10 homes. The data are shown here. Is there a significant difference in the values of the homes for each group? Use a = 0.05.

 Real estate appraisers Tax assessors Mean $83,256$88,354 Standard deviation $3256$2340 Sample size 10 10

Solution: We are interested in testing

\begin{align}{{H}_{0}}:{{\mu }_{1}} {=} {{\mu }_{2}} \\ {{H}_{A}}:{{\mu }_{1}} {\ne} {{\mu }_{2}} \\ \end{align}

which corresponds to a two-tailed independent samples t-test. Before applying the t-test, it is required to test whether the variances can be assumed to be equal or not. We need to test

\begin{align}{{H}_{0}}:\sigma _{1}^{2}=\sigma _{2}^{2} \\{{H}_{A}}:\sigma _{1}^{2}\ne \sigma _{2}^{2} \\ \end{align}

The F-statistics is computed as

$F=\frac{s_{1}^{2}}{s_{2}^{2}}=\frac{{3256}^{2}}{{2340}^{2}}=1.9361$

The lower and upper critical values for $$\alpha =0.05$$ and df1 = 9 and df2 = 9 are

${{F}_{lower}}=0.2484,\,\,\,{{F}_{upper}}=4.026$

which means that we fail to reject the null hypothesis of equal variances. Observe that we are assuming that the variances are equal, so the t-statistics is computed as:

$t=\frac{{{{\bar{X}}}_{1}}-{{{\bar{X}}}_{2}}}{{{s}_{p}}\sqrt{\frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{2}}}}}$

where the pooled standard deviation is computed as

${{s}_{p}}=\sqrt{\frac{\left( {{n}_{1}}-1 \right)s_{1}^{2}+\left( {{n}_{2}}-1 \right)s_{2}^{2}}{{{n}_{1}}+{{n}_{2}}-2}}=\sqrt{ \frac{9\times {3256}^{2}+9 \times {2340}^{2}}{9+9}}= {2835.2369}$

This means that the t-statistics is

$t=\frac{{{{\bar{X}}}_{1}}-{{{\bar{X}}}_{2}}}{{{s}_{p}}\sqrt{\frac{1}{{{n}_{1}}}+\frac{1}{{{n}_{2}}}}}=\frac{{83256}-{88354}}{2835.2369\sqrt{\frac{1}{10}+\frac{1}{10}}}={-4.0206}$

The critical value for $$\alpha = 0.05$$ and for $$df = 18$$ degrees of freedom for this two-tailed test is $$t_{c} = 2.1$$. The rejection region is given by

$R=\left\{ t:\,\,\,|t|>{2.1} \right\}$

Since $$|t| = 4.0206 {>} t_c = 2.1$$, then we reject the null hypothesis H0.

Hence, we have enough evidence to support the claim that there is a significant difference in the values of the homes for each group.

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