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The article also reported that for a sample of 36 individuals who had taken quetiapine, the sample mean cholesterol level change and estimated standard error were 9.06 and 4.256, respectively. Make any necessary assumptions about the distribution of change in cholesterol level, test the hypothesis that the true average cholesterol level increases. Givens: State the appropriate hypotheses. (Use μD = μafter − μbefore.) H0: μD = 0 Ha: μD > 0 t value= 2.1 or 2.128759398 unrounded p value=? Identify the significance levels at which there is sufficient evidence to reject H0. (Choose all that apply.) ? = 0.1 ? = 0.05 ? = 0.01 ? = 0.001 For the sample of 45 individuals who had taken olanzapine, the article reported (7.520, 9.830) as a 95% CI for true average weight gain (kg). What is a 99% CI? (Round your answers to three decimal places.

Sagot :

rafikiedu08

P-value is equal to 0.0206 and the significance levels at which there is sufficient evidence to reject H0 are at = 0.05 and [tex]\alpha[/tex]=0.1  and value of 99% CI = 7.132, 10.218.

Here we have,

n= 36

e= 9.06

S.E. = 4.256

Hypothesis

H0: μD = 0

Ha: μD > 0

This a right-tailed test

Test statistics,

t = 2.12

P-value at t= 2.12 and degree of freedom(df) = n-1 =36-1 = 35

P-value = t-dist(t, df, tails)

            = t-dist(2.12, 35, L)

= 0.020585

= 0.0206 Round to 4  decimals

If pvalue>[tex]\alpha[/tex], then we reject the null hypothesis.

So, [tex]\alpha[/tex]= 0.05 and [tex]\alpha[/tex]=0.1 we reject the H0.

For sample of n=45 df will be= 45 -1 =44

t0.025 = 2.015

Mean weight = (7.520 + 9.830)/2 = 8.675

Margin of error = 9.830 - 8.675 = 1.155

SE = 1.155/2.015 = 0.573

99%tc = 2.692

Margin of error = 2.692 x 0.573 = 1.543

99% CI =  8.675 ± 1.543

           = 7.132, 10.218

99% CI = 7.132, 10.218.

To learn more about the null hypothesis visit: https://brainly.com/question/28920252

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