Question

a certain breed of rat shows a mean weight gain of 65g during the first 3 months of life. a random sample of 34 rats of this breed are fed a new diet from birth until age 3 months. these 34 rats have a mean weight gain of 60.75g and a standard deviation of 3.84g. we are interested in testing whether there is reason to believe, at the 0.05 significance level, that the new diet is causing a change in the average amount of weight gained in this breed of rats. would a 95% confidence interval contain 65, the null value? calculate the confidence interval to verify your decision

Answer 1

The 95% confidence interval for the average weight gain of rats on the new diet is (59.46g, 62.04g), which does not contain the null value of 65g, indicating a potential effect of the diet on weight gain.

Step-by-step explanation:

A 95% confidence interval for the mean weight gain of the rats on the new diet can be computed using the sample mean, the sample standard deviation, and the z-score corresponding to a 95% confidence level for a normal distribution. To check if the confidence interval contains 65g, follow these steps:

• Calculate the standard error (SE) by dividing the sample standard deviation (3.84g) by the square root of the sample size (34).
• Find the z-score corresponding to a 95% confidence level, which is approximately 1.96.
• Calculate the margin of error by multiplying the z-score with the standard error.
• Determine the confidence interval by adding and subtracting the margin of error from the sample mean (60.75g).

If the resulting confidence interval does not include 65, it suggests that the diet has an effect on the average weight gain. Calculating the confidence interval:

1. SE = 3.84 / sqrt(34) = 0.6582
2. Margin of Error = 1.96 * 0.6582 = 1.29g
3. Confidence Interval = 60.75 ± 1.29g = (59.46g, 62.04g)

Since 65g is not within the interval (59.46g, 62.04g), the 95% confidence interval does not contain the null value, suggesting that the new diet has an effect on weight gain. This decision can be made before performing an hypothesis test.

Answer 2

To test whether the new diet is causing a change in the average weight gain of the rats, we can construct a confidence interval. The calculated confidence interval does not contain the null value of 65, indicating that there is reason to believe that the new diet is causing a change in the average weight gain.

Step-by-step explanation:

To test whether there is reason to believe that the new diet is causing a change in the average amount of weight gained in this breed of rats, we can use a confidence interval. Since we are interested in testing at the 0.05 significance level, we can use a 95% confidence level. If the confidence interval contains the null value of 65, then there is no reason to believe that the new diet is causing a change in the average weight gain.

To calculate the confidence interval, we can use the formula:

CI = x ± Z * (σ/√n)

Where:

• x is the sample mean weight gain
• Z is the Z-score corresponding to the desired level of confidence
• σ is the population standard deviation
• n is the sample size

Plugging in the given values, we can calculate the confidence interval:

CI = 60.75 ± 1.96 * (3.84/√34)

CI = 60.75 ± 1.96 * 0.6604

CI ≈ 60.75 ± 1.2941

CI ≈ (59.4559, 62.0441)

Since 65 is not within this confidence interval, we can conclude that there is reason to believe that the new diet is causing a change in the average weight gain.