A manufacturer of candy must monitor the temperature at which the sweets are baked. Too much variation will cause inconsistency in the taste of the sweets. Past records shows that the variance of the temperature has been 1.440. A random sample of 30 batches of sweets is selected and the variance of the temperature is 4.41. Testing at the 1% level of significance whether the population variance differs from the past temperature variance showed the critical value is given as:
In order to determine the critical value for the hypothesis test, we need to find the upper and lower critical values from the chi-square distribution.
Given:
Sample size (n) = 30
Sample variance (s^2) = 4.41
Population variance (σ^2) = 1.44
To test the hypothesis at the 1% level of significance, we use a chi-square distribution with (n-1) degrees of freedom.
The null hypothesis (H0) is that the population variance is equal to the past temperature variance.
The alternative hypothesis (H1) is that the population variance differs from the past temperature variance.
Since we want to test whether the population variance differs, this is a two-tailed test. Therefore, we need to divide the significance level (α) by 2.
Significance level (α) = 0.01
α/2 = 0.01/2 = 0.005 (for each tail)
The degrees of freedom for the chi-square distribution is (n-1).
df = 30 - 1 = 29
Using a chi-square table or calculator, we can find the upper and lower critical values for the 0.005 significance level with 29 degrees of freedom.
The upper critical value (c1) is the value where the cumulative probability is 0.995.
The lower critical value (c2) is the value where the cumulative probability is 0.005.
For a chi-square distribution with 29 degrees of freedom, the upper critical value (c1) is approximately 46.963.
For the lower critical value (c2), we can use the fact that the chi-square distribution is symmetric. So, c2 = -c1 = -46.963.
Therefore, the critical value for this hypothesis test is c1 = 46.963.
To test whether the population variance differs from the past temperature variance, we can use the Chi-Square test for variance.
Step 1: State the null and alternative hypotheses:
- Null Hypothesis (H0): The population variance is equal to the past temperature variance.
- Alternative Hypothesis (Ha): The population variance differs from the past temperature variance.
Step 2: Set the level of significance (α):
The level of significance is given as 1%, which means α = 0.01.
Step 3: Determine the critical value:
To find the critical value for the Chi-Square test, we need the degrees of freedom. In this case, the degrees of freedom (df) is determined by the sample size, which is 30.
Using a Chi-Square distribution table or a statistical software, we find that for a 1% level of significance and 29 degrees of freedom, the critical value is approximately 46.98.
Step 4: Calculate the test statistic:
The test statistic for the Chi-Square test for variance is calculated using the formula:
χ2 = (n-1) * (s^2) / σ^2
Where:
n is the sample size,
s^2 is the sample variance,
and σ^2 is the population variance.
In this case:
n = 30,
s^2 = 4.41,
and σ^2 (past temperature variance) = 1.440.
Plugging in these values, we can calculate the test statistic:
χ2 = (30 - 1) * (4.41) / 1.440 = 86.47
Step 5: Make a decision:
If the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the test statistic (86.47) is greater than the critical value (46.98). Therefore, we reject the null hypothesis.
Step 6: State the conclusion:
Based on the test results, at a 1% level of significance, there is sufficient evidence to conclude that the population variance of the temperature differs from the past temperature variance.