Mathematics, 07.06.2020 05:02 animedragonking459

A researcher uses a matched-samples design to investigate whether single people who own pets are generally happier than single people without pets. A mood inventory questionnaire is administered to a group of 20- to 29-year old non-pet owners and a similar age group of pet owners. The pet owners are matched one-to-one with the non-pet owners for income, number of close friendships and general health. The data are as follows. Matched-Pair Non-pet Pet A 11 13 B 9 8 C 11 14 D 13 13 E 6 12 F 9 11 Calculate and report below the value of the appropriate test statistic to determine if there is evidence for this researcher's hypothesis.

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There is not enough evidence to support the claim that single people who own pets are generally happier than single people without pets. (P-value=0.0509).

Step-by-step explanation:

We have to calculate the difference for every pair, and applied the hypothesis test to this sample of differences.

Non-pet Pet --> Difference (d)

A 11 13 --> 2

B 9 8 --> -1

C 11 14 --> 3

D 13 13 --> 0

E 6 12 --> 6

F 9 11 --> 2

The claim is that single people who own pets are generally happier than single people without pets. In the context of the sample, it means that the difference between the metric of happiness between the pet and the non-pet subjects is bigger than 0.

The mean and standard deviation of the sample of differences is:

$M=\dfrac{1}{6}\sum_{i=1}^{6}(2+(-1)+3+0+6+2)\\\\\\ M=\dfrac{12}{6}=2$

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{6}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{5}\cdot [(2-(2))^2+(-1-(2))^2+(3-(2))^2+(0-(2))^2+(6-(2))^2+(2-(2))^2]}\\\\\\ s=\sqrt{\dfrac{1}{5}\cdot [(0)+(9)+(1)+(4)+(16)+(0)]}\\\\\\ s=\sqrt{\dfrac{30}{5}}=\sqrt{6}\\\\\\s=2.449$

Then we perform an hypothesis test for the population mean.

The claim is that single people who own pets are generally happier than single people without pets.

Then, the null and alternative hypothesis are:

$H_0: \mu_d=0\\\\H_a:\mu_d 0$

The significance level is 0.05.

The sample has a size n=6.

The sample mean is M=2.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.449.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{2.449}{\sqrt{6}}=0.9998$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2-0}{0.9998}=\dfrac{2}{0.9998}=2.0004$

The degrees of freedom for this sample size are:

df=n-1=6-1=5

This test is a right-tailed test, with 5 degrees of freedom and t=2.0004, so the P-value for this test is calculated as (using a t-table):

P-value=P(t2.0004)=0.0509

As the P-value (0.0509) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that single people who own pets are generally happier than single people without pets.

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Answer from: Quest

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A researcher uses a matched-samples design to investigate whether single people who own pets are gen...