This problem has been solved! see the answersuppose we examine the distribution of scores on the last exam separately for freshman and sophomores. both distributions are approximately normally distributed and have the same mean, but the distribution of exam scores had a smaller standard deviation among sophomores than freshman. a z-score is calculated for mary’s exam score relative to both distributions (freshman and sophomores). given that mary’s score is well above the mean, which of the following would be true about these two z-scores? a. the z-score based on the distribution for the freshman students would be higher. b. the z-score based on the distribution for the sophomore students would be higher. c. the two z-scores would be the same. d. there is not enough information to determine

(1) The correct option is (B).

(2) The mean of the distribution of sample means is 19 fl. oz.

(3) The standard deviation of the distribution of sample means is 0.283 fl. oz.

(4) The correct option is (C).

Step-by-step explanation:

According to the Central Limit Theorem if we have a normal population with mean μ and standard deviation σ and a number of random samples are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

The mean of the sampling distribution of sample mean will be:

And the standard deviation of the sampling distribution of sample mean will be:

The information provided is:

μ = 129

σ = 0.80

n = 8

(1)

The shape of the sampling distribution of sample mean will be Normal.

This is because the population from which the sample is selected is normal.

The correct option is (B).

(2)

Compute the mean of the distribution of sample means as follows:

   

Thus, the mean of the distribution of sample means is 19 fl. oz.

(3)

Compute the standard deviation of the distribution of sample means as follows:

   

Thus, the standard deviation of the distribution of sample means is 0.283 fl. oz.

(4)

Any change in the sample size will have no effect on the mean of the distribution of sample means.

But, if the sample is increased or decreased than the standard deviation will be decreased or increased respectively.

This is because the standard deviation of the distribution of sample means is inversely proportional to the sample size.

So, for n = 100 the standard deviation is:

   

Thus, the standard deviation was decreased.

The correct option is (C).

0.229

Step-by-step explanation:

Given that the difference between the two sample means follows anormal distribution with a mean of11.00 and standard deviation equal to1.4387

A statistician is interested in the effectiveness of a weight-loss supplement. She randomly selects two independent samples. Individuals in the first sample of size n1 = 24 take the weight-loss supplement. Individuals in the second sample of size n2 = 21 take a placebo. Individuals in both samples follow identical exercise and diet programs. At the end of the study, the statistician measures the weight loss (in percent) of each participant.

We find that mean difference actual = 13-2 = 11

Probability that difference >12 =P(Z>)

=P(Z>0.742)=.0.229

50, 100, 200 and part 2 is b,c,a

Step-by-step explanation:

e2020

a) True

b) True

c) True

d) False

e) False

Step-by-step explanation:

(a) All other factors remaining the same, increasing the sample size, n, will decrease the width of a confidence interval.

True. Confidence intervals are calculated by calculating margin of error (ME) around the mean using the formula

ME= where

z is the corresponding statistic (z-score or t-score) s is the standard deviation of the sample(or of the population if it is known) N is the sample size

As the formula suggests, all other factors remaining the same, if we increase N, ME decreases.

b) We expect 95% of all 95% confidence intervals for the population mean to contain the sample mean.

True. This is what 95% confidence level assumes.

(c) The t-distribution is symmetric and centered at the population mean

True.

(d) The t-distribution is very similar to the standard normal distribution regardless of its degrees of freedom.

False. As the degrees of freedom increases t-distribution resembles the standard normal distribution. For small sample sizes (

(e) All other factors remaining the same, a 90% confidence interval for a population mean is narrower than an 95% confidence interval for the same population

False. 90% confidence interval for a population mean is wider than an 95% confidence interval for the same population

B is the right answer bro.. The shape is high as high

Step-by-step explanation:

Distribution A : 50

Distribution B : 100

Distribution C : 200

Second part is : B,C,A

Step-by-step explanation: