Question

A survey of 90 men found that an average amount spent on St. Patrick's day of $55 with a standard deviation of $18. A similar survey of 86 women found they spent an average of $44 with a standard deviation of $16. When testing the hypothesis (at the 5% level of significance) that men spend more than women on St. Patrick's day, what is the test statistic?

Answer 1

Answer:

The value of the test statistic is 4.70.

Step-by-step explanation:

The hypothesis for this test can be defined as follows:

H₀: Men do not spend more than women on St. Patrick's day, i.e. μ₁ = μ₂.

Hₐ: Men spend more than women on St. Patrick's day, i.e. μ₁ > μ₂.

The population standard deviations are not known.

So a t-distribution will be used to perform the test.

The test statistic for the test of difference between mean is:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }$

Given:

$\bar x_{1}=55\\s_{1}=18\\n_{1}=90\\\bar x_{1}=44\\s_{1}=16\\n_{1}=86$

Compute the value of the test statistic as follows:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }=\frac{55-44}{\sqrt{\frac{15^{2}}{90}+\frac{16^{2}}{86} }}=4.70$

Thus, the value of the test statistic is 4.70.