What degrees of freedom should be used for the critical t value?

Answer: 15 For a one-sample t-procedure, df = n - 1 = 16 - 1 = 15.

What is the standard error of the mean for this sample?

Answer: 1.02 SE = s/sqrt(n) = 5.1/sqrt(25) = 5.1/5 = 1.02.

What is the minimum sample size needed?

Answer: 62 n = (z sigma/ME)^2 = (1.96 8/2)^2 = (7.84)^2 = 61.5. Round up to 62.

AP Statistics · Topic 7.2

Constructing a Confidence Interval for a Population Mean Practice

Part of Inference for Quantitative Data: Means.(UNC-4.E)

Practice questions

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Sample questions

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Sample 1difficulty 1/5
$A researcher constructs a 95% confidence interval for a population mean from a random sample of size n = 16.$
$What degrees of freedom should be used for the critical t* value?$
- A
  $16$
- B
  $30$
- C
  $14$
- D
  $15$
  check_circle
Why
$For a one-sample t-procedure, df = n - 1 = 16 - 1 = 15.$
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Sample 2difficulty 1/5
$A random sample of 25 batteries has a mean lifetime of 42.3 hours with sample standard deviation 5.1 hours. The lifetimes are approximately normally distributed.$
$What is the standard error of the mean for this sample?$
- A
  $0.20$
- B
  $5.10$
- C
  $1.02$
  check_circle
- D
  $0.41$
Why
$SE = s/sqrt(n) = 5.1/sqrt(25) = 5.1/5 = 1.02.$
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Sample 3difficulty 1/5
$A researcher wants a 95% z-interval with margin of error at most 2 units. From a pilot study, sigma = 8.$
$What is the minimum sample size needed?$
- A
  $16$
- B
  $62$
  check_circle
- C
  $8$
- D
  $32$
Why
$n >= (z* sigma/ME)^2 = (1.96 * 8/2)^2 = (7.84)^2 = 61.5. Round up to 62.$
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Sample 4difficulty 1/5
$A 95% t-interval has the form x-bar \pm t* (s/sqrt(n)). For n = 25, s = 10, x-bar = 50, t* = 2.064.$
$What is the margin of error?$
- A
  $0.83$
- B
  $4.13$
  check_circle
- C
  $20.64$
- D
  $2.06$
Why
$ME = 2.064 * (10/sqrt(25)) = 2.064 * 2 = 4.13.$
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Sample 5difficulty 1/5
$A random sample of n = 20 trees yields a mean diameter of 12.4 inches with s = 2.1 inches. Use t* = 2.093 for 95% confidence with df = 19.$
$What is the 95% confidence interval for the mean tree diameter?$
- A
  $(12.30, 12.50)$
- B
  $(10.30, 14.50)$
- C
  $(11.42, 13.38)$
  check_circle
- D
  $(11.98, 12.82)$
Why
$ME = 2.093 * (2.1/sqrt(20)) = 2.093 * 0.4696 = 0.983. CI = 12.4 \pm 0.98 = (11.42, 13.38).$
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