In estimating the mean score on a fitness exam, we use an original sample of size and a bootstrap distribution containing bootstrap samples to obtain a confidence interval of to . A change in this process is described below. If all else stays the same, which of the following confidence intervals (A, B, or C) is the most likely result after the change: Using an original sample of size n 16?
A. 66 to 74
B. 67 to 73
C. 67.5 to 72.5

Option A is correct.

The confidence interval becomes

66 to 74.

Step-by-step explanation:

Complete Correct Question

In estimating the mean score on a fitness exam, we use an original sample of size n = 30 and a bootstrap distribution containing 5000 bootstrap samples to obtain a 95% confidence interval of 67 to 73. A change in this process is described below. If all else stays the same, which of the following confidence intervals (A, B, or C) is the most likely result after the change: Using an original sample of size n=16?A. 66 to 74

B. 67 to 73

C. 67.5 to 72.

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

To calculate the Sample Mean, we will use the confidence imterval obtained for the first sample size of 30 was 67 to 73.

Let the mean be x and the margin of error = y.

Lower limit of the confidence interval = x - y = 67

Upper limit of the confidence interval = x + y = 73

Solving simultaneously,

Sample mean = 70

Margin of error = 3

Then, since the sample size was initially 30, the critical value was obtained from the z-tables.

95% critical value = z = 1.960

Margin of error = (critical value) × (standard error)

3 = 1.960 × (standard error)

Standard error = 1.5306

Standard error of the mean = σₓ = (σ/√n) = 1.5306

σ = standard deviation of the sample = ?

n = sample size = 30

1.5306 = (σ/√30)

σ = 1.5306 × √30 = 8.3835

Note that using the central limit theorem, we can assume that sample mean for a sample size of 30 people will be approximately equal to the same value, digital value, that is Adam score of 70.

Sample mean for sample size 16, = 70

Sample standard deviation = 8.3835

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

Critical value = 1.960

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 8.3835

n = sample size = 16

σₓ = (8.3835/√16) = 2.096

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 70 ± (1.960 × 2.096)

CI = 70 ± 4.10

95% CI = (65.9, 74.1)

95% Confidence interval = (66, 74)

Hope this Helps!!!

show us the graphs dude

step-by-step explanation:

12cm

the answer is d