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Acrylic bone cement is sometimes used in hip and knee replacements to secure an artificial joint in place. The force required to break an acrylic bone cement bond was measured for seven specimens, and the resulting mean and standard deviation were 306.07 newtons and 41.96 newtons, respectively. Assuming that it is reasonable to believe that breaking force has a distribution that is approximately normal, use a 95% confidence interval to estimate the mean breaking force for acrylic bone cement. (Round your answers to three decimal places.)

Sagot :

Using the t-distribution, it is found that the 95% confidence interval to estimate the mean breaking force for acrylic bone cement, in newtons, is (267.264, 344.876).

We are given the standard deviation for the sample, which is why the t-distribution is used to solve this question.

The information given is:

  • Sample mean of [tex]\overline{x} = 306.07[/tex].
  • Sample standard deviation of [tex]s = 41.96[/tex].
  • Sample size of [tex]n = 7[/tex].

The confidence interval is:

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

The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 7 - 1 = 6 df, is t = 2.4469.

Hence, replacing the values, the interval is:

[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 306.07 - 2.4469\frac{41.96}{\sqrt{7}} = 267.264[/tex]

[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 306.07 + 2.4469\frac{41.96}{\sqrt{7}} = 344.876[/tex]

Then, the 95% confidence interval to estimate the mean breaking force for acrylic bone cement, in newtons, is (267.264, 344.876).

A similar problem is given at https://brainly.com/question/15180581

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