1 an economist is studying salaries for high technology...

An economist is studying salaries for high technology companies and wants to test the claim that the average salary for high tech employees is less than $70,000. The economist selects a sample of 35 random employees from various high tech companies and records their salaries. Based on past studies, the economist determines that the population standard deviation is $8,500. The economist conducts a one-mean hypothesis test at the 5% significance level to test the claim that the average salary for high tech employees is less than $70,000. Which is the correct setup for the null and alternative hypothesis for this example? Select the correct answer below: *A. H0:μ=70,000; Ha:μ<70,000, which is a left-tailed test. B. H0:μ=70,000; Ha:μ>70,000, which is a right-tailed test C. H0:μ<70,000; Ha:μ=70,000, which is a left-tailed test. D. H0:μ>70,000; Ha:μ=70,000, which is a left-tailed test.

Transcript

00:01 So we have a personal director in a particular state claims that the mean annual income is greater in one of the state's counties than is another county, specifically a is greater than b.

00:10 So this person's hypothesis, what this director is claiming is that a is greater than b.

00:18 So that's actually our alternative hypothesis that the mean income, and we're looking at the mean annual income of a is greater than that of b.

00:27 That means the alternative hypothesis would be that, you know what, they're equal.

00:30 You could even say that b could be greater, but the most important part is this alternative, that a is greater than b, the mean of a is greater than b.

00:40 And we're gonna test this claim at the alpha of 0 .05 level of significance.

00:46 And we're given some information.

00:48 There was a sample of size n equals 17 residents in county a, and then the mean of that sample is $41 ,800, same deviation of $8 ,900.

01:02 In county b, there were eight residents, $37 ,900 was that sample mean, and 5 ,400 was the same deviation of that sample.

01:11 And we're gonna assume that the population variances are not equal.

01:14 So we're gonna assume that the variances are not equal.

01:19 So, but we are gonna assume that these are normally distributed.

01:23 So we're gonna, you can use our t tables.

01:26 So it's a two sample t test.

01:31 Sample t test with unequal variances.

01:41 So let's go ahead and get our critical value because we're gonna use that critical value approach to test the algorithm, to make our decision.

01:47 So our rule is to reject h naught.

01:53 If the, if our test statistic is less than some negative t critical value, actually, no, sorry, it's all, sorry.

02:05 If our test statistic is greater than some t critical value and the critical value is based on this alpha.

02:15 It's a one -way test.

02:16 So we're putting all this alpha in one tail.

02:18 So the critical value is based on that alpha, 0 .05.

02:24 And then we have our degrees of freedom.

02:27 We have to do a little work here to get our degrees of freedom.

02:30 It's called, and this is what we do when we have unequal variances.

02:33 It's called welch's degrees of freedom.

02:35 And this is the formula we use.

02:38 So s1 squared, s2 squared, those are the sample variances.

02:42 So we take the standard deviations and square them and that's how we get the variances...

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