Some Examples: For each of the following scenarios, set up the null and alternat

Question

Some Examples: For each of the following scenarios, set up the null and alternative hypothesis, calculate the value of the test statistic t and shade the area under the t-density that corresponds to the p-value. Example 1: Developing a new diet (to lose weight) and measure the # of lbs lost. A random sample of 36 people on the diet yielded an average # of pounds lost of 2.9 lbs with a standard deviation of 5.7 lbs. Is there sufficient evidence in the data to conclude that the diet is effective? Ho versus P-value: t 226 Module 6, Chapter 16 25/47 Example 2 A random sample of 30 Stat 226 students yielded an average number 1.86 caffeinated drinks per day and a standard deviation of 1.28. Is there evidence that the average number of caffeinated drinks per day for all Stat 226 students is different from 1.57 versus p-value:

Explanation / Answer

1. n = 36
Mean = 2.9
Stdev = 5.7

Ho: Diet is not effective, Diff < 2.9
Ha: Diet is effective, Diff >=2.9

t* = (-2.9)/(5.7/sqrt(36)) = -3.05
At df of n-1=35 and t* = -3.05

The P-Value is .002171. The result is significant at p < .05.

There is sufficient eveidence to conclude that diet is effective

2.

Ho: Diff = 1.5

Ha: Diff != 1.5

n=30
Sample.Mean = 1.86
Stdev = 1.28
pop.mean = 1.5

t* = (1.86-1.5)/(1.28/sqrt(30)) = 1.541

At a df of n-1 = 30-1 = 29 and t* = 1.541 we have:

The P-Value is .134402.

The result is not significant at p < .05.

There is not enought evidence to prove that the difference is not equal from 1.5

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