Consider the following hypothesis test: H0 : ? = 110 Ha : ?...

Consider the following hypothesis test: H0 : ? = 110 Ha : ? ? 110 A sample of 85 is used. Identify the p-value and state your conclusion for each of the following sample results. Use ? = 0.02. Round the value of the test statistic to three decimal places. If required enter negative values as negative numbers. a. x? = 114 and s = 10.5 t-value 3.51 p-value lower than .01 conclusion reject H0 b. x? = 104 and s = 11.5 t-value -.57 p-value lower than .01 conclusion reject H0 c. x? = 113 and s = 12.5 t-value 2.18 p-value between .025 and .05 conclusion do not reject H0

Transcript

00:01 So we're looking at hypothesis tests.

00:03 Now the null hypothesis is that the population mean is 110.

00:07 The alternative is it isn't.

00:09 It's something else.

00:11 So all of these are going to be two -tailed tests.

00:15 We need to keep that in mind.

00:18 Okay, the sample size for all of these is 85.

00:21 The level of significance, alpha, is 0 .02.

00:25 So i'm going to draw a t -distribution with 84 degrees of freedom.

00:30 Them.

00:31 T distributions just look, they look like the normal distribution but with wider tails.

00:35 Two tails means we have two rejection regions.

00:39 They split alpha equally between them.

00:42 So each is 1%.

00:43 We have t and minus t, the other critical values, with the other 98 % between them.

00:53 Okay, so so in a, the sample mean is 114, the sample standard deviation is 10 .5.

01:04 So you have got the calculations correct for your t -value, i'm pretty sure.

01:09 The problem is you haven't rounded them to three decimal places.

01:13 But let's go through this.

01:16 The test statistic is basically the z -score of your sample mean.

01:21 And a z -score tells you how many standard deviations away from a mean a value is.

01:27 So this distribution, the sample means for an approximately normal distribution mean same as population mean, standard deviation sigma over root n.

01:37 Sigma is the population standard deviation.

01:39 We don't know what it is.

01:40 That's why we're using t instead.

01:42 It accounts for having to use the sample standard deviation.

01:46 So you take your raw value, the sample mean, subtract the mean, which we think is 110, and divide by what we think is the standard deviation.

01:55 So in the first part, i have 114 minus 110, divided by 10 .5 over root 85, which is 3 .512.

02:11 Now i need to take that and get the p -value.

02:13 So the p -value is the probability of getting this result if the null hypothesis is true.

02:20 So we've got these rejection regions.

02:22 The p -value is the probability of getting something this high.

02:26 We also have the opposite tail as well.

02:30 So area to the right of that and to the left of minus 3 .512.

02:35 I'm going to get this using my calculator with the tcdf function...

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