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Sagot :
Using the t-distribution, it is found that the 95% t-confidence interval for the population mean weight of newborn mice with snipped tails is (42157, 42843).
The interpretation is: The researchers are 95% confident that the true population mean weight of newborn mice with snipped tails is between the upper and lower limits of the confidence interval.
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} = 42500[/tex].
Sample standard deviation of [tex]s = 6068[/tex].
Sample size of [tex]n = 1200[/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 1200 - 1 = 1199 df, is t = 1.96.
The interval is:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 42500 - 1.96\frac{6068}{\sqrt{1200}} = 42157[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 42500 + 1.96\frac{6068}{\sqrt{1200}} = 42843[/tex]
The 95% t-confidence interval for the population mean weight of newborn mice with snipped tails is (42157, 42843).
The interpretation is that we are 95% sure that the population mean weight is in this interval, hence:
The researchers are 95% confident that the true population mean weight of newborn mice with snipped tails is between the upper and lower limits of the confidence interval.
A similar problem is given at https://brainly.com/question/25417022
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