Consider the following hypotheses h0 p 051 ha p 051 which of...

Consider the following hypotheses: H0: p ≥ 0.51 HA: p < 0.51 Which of the following sample information enables us to reject the null hypothesis at α = 0.01 and at α = 0.05? (You may find it useful to reference the appropriate table: z table or t table) (Round all intermediate calculations to at least 4 decimal places.) α = 0.01 α = 0.05 a. x = 47; n = 112 Do not reject H0 Reject H0 Do not reject H0 Reject H0 b. x = 124; n = 326 Reject H0 Do not reject H0 Reject H0 Do not reject H0 c. p̄ = 0.46; n = 56 Do not reject H0 Reject H0 Do not reject H0 Reject H0 d. p̄ = 0.46; n = 446 Do not reject H0 Reject H0 Do not reject H0 Reject H0

Transcript

00:01 So for this problem, what i'll do is first just go over the sort of approach that we'll use, and then i'll show the actual calculations using a ti84, sort of automating it to a degree.

00:11 So we have our null hypothesis is that p is greater than or equal to 0 .51, and our alternate hypothesis is that p is less than 0 .51.

00:21 The way that we would go about calculating this is we would find the test statistic, zd not, which is equal to the sample proportion p hat minus it would be our population or our hypothesized population proportion 0 .51 then we divide that by the square root of 0 .51 times 1 minus 0 .51 divided by the sample size n where it looks like for each one of these we're getting an x and an n value so it would be x over n minus 0 .51 etc.

00:59 So, that being said, we then would have that the p value, or actually i'll say it this way, we would reject the null hypothesis if the p value we calculate, which is equal to the probability of a z score being less than our standardized test statistic, z -not, is less than our alpha value.

01:31 So, in that case, let's see here.

01:34 I'll go stat, tests, z test, stats.

01:42 The mu not value that i'm putting in is 0 .51.

01:45 The standard deviation, or sigma, is going to be calculated by taking the square root of 0 .51 times 0 .49, divided by for the first case, we have n equals 112, so divided by 112.

02:07 X bar is going to be 47 over 112, and n is equal to 112.

02:17 We're testing alternate hypothesis, mu, less than mu not.

02:21 So for part a, we find that the probability is equal to, oh, pardon me, silly mistake.

02:30 I should have been using the prop z test here.

02:32 So actually, p .0 .0 .01.

02:35 X for the first one is 47, n is 112.

02:43 Calculate that.

02:44 We find our p values equal to 0 .028 roughly.

02:48 So we reject at alpha equals 0 .05.

02:55 We fail to reject at alpha equals 0 .01.

03:02 And so what we'll be doing is sort of tabulating the results for each part of the problem in the table that i have here showing the z, the p value, as well as our conclusions at the different values.

03:13 So we'll go through and do our one proportion test again.

03:20 So we have still p -not is 0 .51 for part b.

03:24 Now we have for part b, x is 124.

03:29 N is 326...

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