Finding the critical value(s) of x^2, and use a x^2 distribution table to find the limits containing the P-value. Determine whether there is sufficient evidence to support the given alternative hypothesis.

Test Ho:theta=2.63,Versus H1:theta does not equal2.62, given that theta = 0.10, n=13, and s=2.66.

To find the critical value(s) and use a chi-square (x^2) distribution table to find the limits containing the p-value, we need to follow these steps:

Step 1: Determine the degrees of freedom (df).
The degrees of freedom for a chi-square test is calculated by df = n - 1, where n is the sample size. In this case, n = 13, so df = 13 - 1 = 12.

Step 2: Determine the critical value(s).
To find the critical value(s), we need to determine the level of significance (α), which indicates the probability of making a Type I error. Let's assume α = 0.05 for a 95% confidence level. You can adjust this value based on the desired confidence level.

Looking up the chi-square distribution table for df = 12 and α = 0.05, we find the critical values as follows:

- The lower critical value is denoted as x1 and represents the chi-square value below which we reject the null hypothesis.
- The upper critical value is denoted as x2 and represents the chi-square value above which we reject the null hypothesis.

In this case, x1 = 3.815 and x2 = 21.026.

Step 3: Calculate the test statistic.
The test statistic is given by the formula: χ^2 = ((n-1) * s^2) / σ^2.

In this case, the given sample mean (θ) is 2.63, and we want to determine if it is significantly different from 2.62. Therefore, we can calculate the test statistic as follows:

χ^2 = ((n-1) * s^2) / σ^2
= ((13-1) * 2.66^2) / (2.62^2)
≈ 20.5

Step 4: Compare the test statistic with the critical values.
If the test statistic falls within the range of critical values (x1 and x2), we fail to reject the null hypothesis. If it falls outside this range, we reject the null hypothesis and there is sufficient evidence to support the alternative hypothesis.

In this case, the test statistic (approximately 20.5) is greater than x2 (21.026). Since it falls outside the range of critical values, we reject the null hypothesis and there is sufficient evidence to support the alternative hypothesis.

Therefore, based on the given data, with a significance level of 0.05, we have enough evidence to reject Ho: θ = 2.63 and conclude that θ is not equal to 2.62.