The manager of a service station is in the process of analyzing the number of times car owners change the oil in their cars. She believes that the average motorist changes his or her car's oil less frequently than recommended by the owner's manual (two times per year). In a preliminary survey she asked 15 cars owners how many times they changed their car's oil in the last 12 months. The results are listed below.

1 1 2 0 3
3 0 1 0 1
2 3 1 3 1

The value of the test statistic in this problem is approximately equal to_____?

What would be your decision if a hypothesis test was conducted on this problem with the null hypothesis given as H0 : µ ≥ 2 and the alternate hypothesis given as H1 < 2 ?

1) Reject H0 at the 10%, 5% and 1% level of significance.

2) Reject H0 at the 10% and 5% level of significance but do not reject H0 at the 1% level of significance.

3) Reject H0 at the 10% level of significance but do not reject H0 at the 5% or 1% level of significance.

4) Do not reject H0 at either the 10%, 5% or 1% level of significance

Z-0.0332

0.0332

t value = –2.6458

To answer the first question about the value of the test statistic, we need to calculate it using the given data. The test statistic used in this case is the t-statistic, which helps determine if the sample mean is significantly different from a hypothesized population mean.

To calculate the t-statistic, we need to use the formula:

t = (sample mean - hypothesized mean) / (sample standard deviation / √sample size)

Let's calculate it step by step:

1. Calculate the sample mean:
Add up all the values and divide by the sample size (15 in this case):
(1 + 1 + 2 + 0 + 3 + 3 + 0 + 1 + 0 + 1 + 2 + 3 + 1 + 3 + 1) / 15 = 23 / 15 = 1.5333 (rounded to four decimal places)

2. Calculate the sample standard deviation:
Subtract the sample mean from each value, square the results, sum them up, divide by (sample size - 1), and take the square root:
√[(1 - 1.5333)^2 + (1 - 1.5333)^2 + (2 - 1.5333)^2 + (0 - 1.5333)^2 + (3 - 1.5333)^2 + (3 - 1.5333)^2 + (0 - 1.5333)^2 + (1 - 1.5333)^2 + (0 - 1.5333)^2 + (1 - 1.5333)^2 + (2 - 1.5333)^2 + (3 - 1.5333)^2 + (1 - 1.5333)^2 + (3 - 1.5333)^2 + (1 - 1.5333)^2] / (15 - 1) = √(5.7333 / 14) = √0.40952 (rounded to five decimal places) = 0.63928 (rounded to five decimal places)

3. Calculate the t-statistic:
Use the formula and substitute the values:
t = (1.5333 - 2) / (0.63928 / √15) = -0.4667 / (0.63928 / √15) = -0.4667 / 0.16515 = -2.8236 (rounded to four decimal places)

Therefore, the value of the test statistic is approximately -2.8236.

Moving on to the second question about the decision made in a hypothesis test, we compare the obtained test statistic to critical values from the t-distribution based on the given significance levels (10%, 5%, and 1%).

In this case, the null hypothesis (H0) states that the population mean (µ) is greater than or equal to 2, and the alternative hypothesis (H1) states that the population mean is less than 2.

To make a decision, we compare the test statistic (-2.8236) to the critical values from the t-distribution table for the given significance levels.

Looking at the options:

1) Reject H0 at the 10%, 5%, and 1% level of significance.
2) Reject H0 at the 10% and 5% level of significance but do not reject H0 at the 1% level of significance.
3) Reject H0 at the 10% level of significance but do not reject H0 at the 5% or 1% level of significance.
4) Do not reject H0 at either the 10%, 5%, or 1% level of significance.

For each option, we compare the test statistic (-2.8236) to the corresponding critical value at each significance level. If the test statistic falls beyond the critical value, we reject the null hypothesis; otherwise, we fail to reject it.

Since the test statistic (-2.8236) does not fall beyond the critical values for any of the significance levels, the correct decision is:

4) Do not reject H0 at either the 10%, 5%, or 1% level of significance.