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Sagot :
Using the z-distribution, it is found that the 90% confidence interval for mean calories in a 30-gram serving of all chocolate chip cookies is (143, 149).
We are given the standard deviation for the population, which is why the z-distribution is used to solve this question.
The information given is:
- Sample mean of [tex]\overline{x} = 146[/tex].
- Population standard deviation of [tex]\sigma = 12[/tex].
- Sample size of [tex]n = 65[/tex].
The confidence interval is:
[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]
The critical value, using a z-distribution calculator, for a 95% confidence interval is z = 1.645, hence:
[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 146 - 1.645\frac{12}{\sqrt{35}} = 143[/tex]
[tex]\overline{x} + z\frac{\sigma}{\sqrt{n}} = 146 + 1.645\frac{12}{\sqrt{35}} = 149[/tex]
The 90% confidence interval for mean calories in a 30-gram serving of all chocolate chip cookies is (143, 149).
A similar problem is given at https://brainly.com/question/16807970
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