Solution
If many samples of size "n_i" are taken from a population with mean "\\mu" and standard deviation "\\sigma", the distribution of sample means "\\bar x_i" will have a mean "\\mu" and a standard deviation of "\\frac{\\sigma}{\\sqrt{n_i}}"
Therefore: "\\bar x" ~ "N( 1, 2)"
The test statistic
"t = \\frac{\\bar x-\\mu}{\\frac{\\sigma}{\\sqrt{n_i}}}"At the level of significance, "\\alpha = 0.1" (since this is a one tail test) and degrees of freedom "v=\\ n_i - 1 = 24" , the test statistic
"t = 1.318"
Since we are testing the alternative hypothesis that "\\mu < 1" ,
Aminah rejects the null hypothesis if "\\bar x < -1.636"
When the sample mean "\\bar x = -2"
when "t = 1.5", and degrees of freedom "v=24" the p-value is "= 0.0733"
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