It is claimed that an automobile is driven on the average more than 20,000 kilometers per year. Random sample of 10 automobile owners are asked to keep a record of the kilometers they travel. If the sample mean and standard deviation of automobile driven are found to be 23,500 kilograms and 3900 kilometers respectively, assuming that the kilometers they travel is normally distributed. Test this claim at the 0.05 level of significance.
To test this claim at the 0.05 level of significance, we will perform a hypothesis test using the given sample mean and standard deviation.
Here, we have the following information:
Population mean (claimed): μ = 20,000 kilometers
Sample mean: x̄ = 23,500 kilometers
Sample standard deviation: σ = 3,900 kilometers
Sample size: n = 10
We will set up the null and alternative hypotheses:
Null hypothesis (H₀): The population mean is less than or equal to 20,000 kilometers. (μ ≤ 20,000)
Alternative hypothesis (H₁): The population mean is greater than 20,000 kilometers. (μ > 20,000)
To test this claim, we will use a one-sample t-test because we have a sample mean and the population standard deviation is unknown.
The test statistic for the one-sample t-test is calculated using the formula:
t = (x̄ - μ) / (σ / √n)
where:
- x̄ is the sample mean
- μ is the population mean
- σ is the population standard deviation
- n is the sample size
Plugging in the values, we get:
t = (23,500 - 20,000) / (3,900 / √10)
Calculating this, we find the value of t to be approximately 2.94.
Next, we need to determine the critical value for a one-tailed test with a significance level of 0.05. Since the alternative hypothesis states that the population mean is greater than 20,000, we are performing a right-tailed test.
Looking up the critical value for a one-tailed test with significance level 0.05 and degrees of freedom (df = n - 1 = 10 - 1 = 9) in the t-table, we find the critical value to be approximately 1.833.
Now, we compare the calculated t-value with the critical value.
- If the calculated t-value is greater than the critical value, we can reject the null hypothesis.
- If the calculated t-value is less than or equal to the critical value, we fail to reject the null hypothesis.
In this case, the calculated t-value (2.94) is greater than the critical value (1.833). Therefore, we can reject the null hypothesis.
Thus, there is evidence to support the claim that automobiles are driven on average more than 20,000 kilometers per year.