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The length of time it takes a worker to assemble a large product is assumed to be normally distributed. a random sample of 25 workers a were selected, and their times to assemble the product were recorded. the mean and variance for this sample were 3 and 0.5 hours, respectively.

the 90 percent confidence interval for the mean length of time (in hours) it takes a worker to assemble the product _____ hours to ______ hours

Answer :

Using the t-distribution, it is found that the 90% confidence interval is of  2.71 hours to 3.29 hours.

What is a t-distribution confidence interval?

The confidence interval is:

[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]

In which:

  • [tex]\overline{x}[/tex] is the sample mean.
  • t is the critical value.
  • n is the sample size.
  • s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 25 - 1 = 24 df, is t = 2.0639.

The other parameters are given as follows:

[tex]\overline{x} = 3, s = \sqrt{0.5} = 0.707, n = 25[/tex]

Hence the bounds of the interval are given as follows:

[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 3 - 2.0639\frac{0.707}{\sqrt{25}} = 2.71[/tex]

[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 3 + 2.0639\frac{0.707}{\sqrt{25}} = 3.29[/tex]

More can be learned about the t-distribution at https://brainly.com/question/16162795

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