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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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