Question

A team averaging 110 points is likely to do very well during the regular season. The coach of your team has hypothesized that your team scored at an average of less than 110 points in the years 2013-2015. Test this claim at a 1% level of significance. For this test, assume that the population standard deviation for relative skill level is unknown.
1. Calculate and print the mean points scored by your team during the years you picked.
2. Identify the mean score under the null hypothesis. You only have to identify this value and do not have to print it.
3. Assuming that the population standard deviation is unknown, use Python methods to carry out the hypothesis test.
4. Calculate and print the test statistic rounded to two decimal places.
5. Calculate and print the P-value rounded to four decimal places.

Answer 1

Answer:

1. M=108

2. μ=110

3. In the explanation.

4. Test statistic t = -1.05

5. P-value = 0.1597

Step-by-step explanation:

The question is incomplete: to solve this problem, we need the sample information: size, mean and standard deviation.

We will assume a sample size of 10 matches, a sample mean of 108 points and a sample standard deviation of 6 points.

1. The mean points is the sample points and has a value of 108 points.

2. The null hypothesis is H0: μ=110, meaning that the mean score is not significantly less from 110 points.

3. This is a hypothesis test for the population mean.

The claim is that the mean score is significantly less than 110.

Then, the null and alternative hypothesis are:

$H_0: \mu=110\\\\H_a:\mu< 110$

The significance level is 0.05.

The sample has a size n=10.

The sample mean is M=108.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6}{\sqrt{10}}=1.9$

4. Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{108-110}{1.9}=\dfrac{-2}{1.9}=-1.05$

The degrees of freedom for this sample size are:

$df=n-1=10-1=9$

5. This test is a left-tailed test, with 9 degrees of freedom and t=-1.05, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t<-1.05)=0.1597$

As the P-value (0.1597) 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 the mean score is significantly less than 110.