in a sample of 40 adults on a popular weight loss program, the mean weight loss was 2.1 pounds with a standard deviation of 4.8 pounds. construct a 90% confidence interval estimate of the population mean weight loss for this diet

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(0.82 , 3.38) is estimate of the population mean weight loss for this diet in standard deviation.

What is standard deviation in math?

  • Your dataset's average level of variability is represented by the standard deviation.
  • It reveals the average deviation of each statistic from the mean.
  • A low standard deviation denotes that values are grouped close to the mean, whereas a large standard deviation shows that values are often far from the mean.

mean weight loss = 2.1lb

standard deviation = 4.8lb

sample size n = 40

std error = stdev/sqrt(n) = 4.8/sqrt(40) = 0.76

degrees of freedom = n - 1 = 39

critical value for alpha 0.10 and df 39 = +1.68

margin of error = critical value * std error = +1.68 * 0.76 = +1.28

confidence interval = sample mean + margin of error = 2.1 +1.28 = (0.82 , 3.38)

Yes it is effective since the values > 0mean weight loss = 2.1lb

standard deviation = 4.8lb

sample size n = 40

std error = stdev/sqrt(n) = 4.8/sqrt(40) = 0.76

degrees of freedom = n - 1 = 39

critical value for alpha 0.10 and df 39 = +1.68

margin of error = critical value * std error = +1.68 * 0.76 = +1.28

confidence interval = sample mean + margin of error = 2.1 +1.28 = (0.82 , 3.38)

Yes it is effective since the values > 0

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