Question

Using the table below, would you please check my answers. The answer for the p-value is supposed to be 0.0000; however, I keep getting 0.9999999977 as the answer. How do you find the correct p-value? Where am I going wrong in my calculation?

Regardless of your answer to part (a) using R, run a t-test for the difference between the two treatment means at the α = 5% level of significance. Specifically, we are interested in knowing if there exists sufficient statistical evidence to support the claim that low temperatures increase blood pressure in rats. Be sure to state the null and alternative hypotheses, calculate the appropriate test statistic, calculate the p-value, and state your conclusions.

The null hypothesis: H₀: µ₁= µ₂(the blood pressures are equal)

The alternative hypothesis: H_α: µ₁> µ₂(blood pressure is less at a lower temperature)

Welch Two Sample t-test

data: bloodpressure5 and bloodpressure26
t = 18.509, df = 8.3215, p-value = 2.348e-08
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
191.4077 Inf
sample estimates:
mean of x mean of y
378.5000 165.8333

standard deviation (bloodpressure26) 14.77047

standard deviation(bloodpressure5)  23.95621

Test statistic = -18.509
critical t-value (using the t-table for right-tailed probabilities) =1.812

right-tailed p-value = (-18.5095, 10) = 0.9999999977 ????

Answer 1

Note- In the question, we are interested in knowing if there exists sufficient statistical evidence to support the claim that low temperatures increase blood pressure in rats.

So, in alternative hypothesis we have to consider blood pressure is more at a lower temperature. But in your solution you considered alternative hypothesis as blood pressure is less at a lower temperature. So, the alternative hypothesis which was considered in your solution is incorrect.

Solution-

The null hypothesis

The alternative hypothesis

So, it is a left tailed (less than type) test.

Test statistic = -18.509

Critical t-value = -1.812    [Using R-code 'qt(0.05,10)']

   [Using R-code 'pt(-18.509,10)']