Answer:
Reject the null hypothesis. There is sufficient evidence to conclude that the mean is less than 31%.
Step-by-step explanation:
In this case we need to test whether the popular charity has expenses that are higher than other similar charities.
The hypothesis for the test can be defined as follows:
<em>H</em>₀: The popular charity has expenses that are higher than other similar charities, i.e. <em>μ </em>> 0.31.
<em>Hₐ</em>: The popular charity has expenses that are less than other similar charities, i.e. <em>μ </em>< 0.31.
As the population standard deviation is not known we will use a t-test for single mean.
Compute the sample mean and standard deviation as follows:
![\bar x=\frac{1}{n}\sum X=\frac{1}{10}\cdot[0.26+0.12+...+0.26]=0.238\\\\s= \sqrt{ \frac{ \sum{\left(x_i - \overline{X}\right)^2 }}{n-1} } = \sqrt{ \frac{ 0.0674 }{ 10 - 1} } =0.08654\approx 0.087](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7Bn%7D%5Csum%20X%3D%5Cfrac%7B1%7D%7B10%7D%5Ccdot%5B0.26%2B0.12%2B...%2B0.26%5D%3D0.238%5C%5C%5C%5Cs%3D%20%5Csqrt%7B%20%5Cfrac%7B%20%5Csum%7B%5Cleft%28x_i%20-%20%5Coverline%7BX%7D%5Cright%29%5E2%20%7D%7D%7Bn-1%7D%20%7D%09%09%09%09%09%09%20%3D%20%5Csqrt%7B%20%5Cfrac%7B%200.0674%20%7D%7B%2010%20-%201%7D%20%7D%20%3D0.08654%5Capprox%200.087)
Compute the test statistic value as follows:

Thus, the test statistic value is -2.62.
Compute the p-value of the test as follows:


*Use a t-table.
Thus, the p-value of the test is 0.014.
Decision rule:
If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.
p-value = 0.014 < α = 0.05
The null hypothesis will be rejected at 5% level of significance.
Thus, concluding that there is sufficient evidence to conclude that the mean is less than 31%.