Chocolate Chip Cookie Calories The average 1-ounce chocolate chip cookie contains 110 calories. A random sample of 13 different brands of 1-ounce chocolate chip cookies resulted in the following calorie amounts. At the =a0.01 level, is there sufficient evidence that the average calorie content is greater than 110 calories? Assume the population is normally distributed. 105 100 170 100 110 115 150 120 165 120 125 153 100 Find the critical value(s). Round the answer to three decimal places, if necessary. If there is more than one critical value, separate them with commas. Critical value(s):
The Correct Answer and Explanation is :
To determine whether the average calorie content of chocolate chip cookies is greater than 110 calories, we will perform a one-sample t-test. Let’s go through the steps of the hypothesis test and calculate the critical value.
Step 1: State the hypotheses
We are testing if the average calorie content is greater than 110 calories. Therefore, our hypotheses are:
- Null hypothesis ((H_0)): The mean calorie content is 110 calories ((\mu = 110)).
- Alternative hypothesis ((H_1)): The mean calorie content is greater than 110 calories ((\mu > 110)).
Step 2: Identify the significance level and test statistic
We are using a significance level of ( \alpha = 0.01 ), which means we will reject the null hypothesis if the test statistic exceeds the critical value at this level.
The test statistic formula for a one-sample t-test is:
[
t = \frac{\bar{x} – \mu}{\frac{s}{\sqrt{n}}}
]
where:
- (\bar{x}) is the sample mean,
- (\mu) is the population mean (110 calories),
- (s) is the sample standard deviation,
- (n) is the sample size.
Step 3: Calculate the sample mean and standard deviation
We are given the following sample data:
105, 100, 170, 100, 110, 115, 150, 120, 165, 120, 125, 153, 100.
Let’s calculate the sample mean ((\bar{x})) and sample standard deviation ((s)).
Step 4: Find the critical value for ( \alpha = 0.01 )
Since we are conducting a one-tailed test with (\alpha = 0.01), we will use a t-distribution with 12 degrees of freedom (since (n = 13), so (df = n – 1 = 12)) to find the critical value.
Let me calculate these values for you.
The sample mean ((\bar{x})) is approximately 125.62, and the sample standard deviation (s) is approximately 25.30.
For a one-tailed t-test with a significance level of (\alpha = 0.01) and 12 degrees of freedom, the critical value is approximately 2.681.
Explanation:
- The critical value tells us the threshold beyond which we would reject the null hypothesis. Since we are conducting a one-tailed test (because we’re testing if the mean is greater than 110 calories), we compare the calculated t-statistic to this critical value.
- If the calculated t-statistic exceeds the critical value of 2.681, we reject the null hypothesis and conclude that there is sufficient evidence that the average calorie content is greater than 110 calories.
Let me know if you want me to continue with the complete test or create any visualizations!