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A safety auditor at a paint factory took a random sample of 25 cans of paint from a production run of over
1,000 cans and measured the amount of lead in each can of paint in the sample. The sample data were roughly
symmetric with a mean of 37 parts per million (ppm) of lead per can and a standard deviation of 18 ppm.
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As
Based on this sample, which of the following is a 95% confidence interval for the mean amount of lead per can
of paint (in ppm) in this production run?

Respuesta :

Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval is (29.6, 44.4).

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 by:

[tex]\overline{x} = 37, s = 18, n = 25[/tex].

Hence, the bounds of the interval are given as follows:

[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 37 - 2.0639\frac{18}{5} = 29.6[/tex]

[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 37 + 2.0639\frac{18}{5} = 44.4[/tex]

The 95% confidence interval is (29.6, 44.4).

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

my1stcatsnameisfrost

Answer:

To simplify what the other person said:

(29.57, 44.43)

Step-by-step explanation:

Proof that this is right on Khan

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