Answer:
The p-value of the test is 0.0613 > 0.05, which means that there has not been a statistically significant increase in the number of attempts.
Explanation:
Typically there are about 52 such attempts noted each day. Test if there has been a statistically significant increase.
At the null hypothesis, we test if the mean is the same, that is of 52. Thus:
At the alternative hypothesis, we test if there has been an increase, that is, if the mean is greater than 52. So
Use the standard level for alpha.
So
The test statistic is:
We have the standard deviation for the sample, so t-distribution.
In which X is the sample mean,
is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.
52 is tested at the null hypothesis:
This means that
Over the previous 30 days in which there were an average of 57 attempts each day with a standard deviation of 17.2.
This means that
Value of the test statistic:
P-value of the test and decision:
The p-value of the test is the probability of finding a sample mean above 57, which is a right-tailed test with t = 1.59 and 30 - 1 = 29 degrees of freedom.
Using a t-distribution calculator, this p-value is of 0.0613.
The p-value of the test is 0.0613 > 0.05, which means that there has not been a statistically significant increase in the number of attempts.