Question

suppose that a restaurant chain claims that its bottles of ketchup contain 24 ounces of ketchup on average, with a standard deviation of 0.2 ounces if you took a sample of 49 bottles of ketchup, what would be the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample?

Answer 1

The answer is 24 +- 0.057

Explanation:

s the sample size. In this case, the sample mean is 24 ounces, the population standard deviation is 0.2 ounces, and the sample size is 49. The critical value for a 95% confidence level is approximately 1.96. Plugging these values into the formula gives us: 24pm1.96frac0.2sqrt49=24pm0.056

So, the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample is (23.944, 24.056).

Answer 2

The 96% confidence interval for the mean number of ounces of ketchup per bottle in the sample is 24 ± 0.029

Explanation:

Given

Mean, x = 24 ounces

Standard deviation, σ = 0.2 ounces

Sample Size, n = 49 bottles of ketchup

Confidence Interval of 95% (0.95)

To calculate the confidence at this level, the following steps are to be followed;

1. Calculating degree of freedom:

Degree of freedom (df) is calculated by subtracting 1 from the sample size.

df = n - 1

df = 49 - 1

df = 48

2. Subtract the confidence level from 1, then divide by two.

α = (1 – .95) / 2 = 0.025

3. Look up the answers to (1) and (2) in the t-distribution table.

For 48 degrees of freedom (df) and α = 0.025,

We get 2.0106

4. Divide sample standard deviation by the square root of sample size.

i.e σ/√n

= 0.2/√49

= 0.2/7

= 0.02857

= 0.029

5. Finally, the 95% confidence interval is calculated using the following illustration.

X ± result in step 5

Where x = 24

So, we have

24 ± 0.029