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 ```I have some statistical questions that I need answered. I would mainly like to know how to solve the critical z score value. As well as some background on the concepts related to how it was solved. These are the following questions: AA random sample of 100 observations from a population with standard deviation 22.5494377436275 yielded a sample mean of 94.4. 1. Given that the null hypothesis is u=90 and the alternative hypothesis is u>90 using a (alpha)=.05, find the following: critical z score test statistic 2. If the null hypothesis is u =90, alt hypothesis u [not =] 90, using a (alpha) = .05 find the following: positive crit. z score, neg. z score, test statatistic Thanks, J```
 ```Hello! Here are your answers. Question 1 The procedure to solve questions regarding hypothesis testing on the population mean is the following. First of all, you need to identify whether you're using a one- or a two-tailed test. This depends on the form of the alternative hypothesis: for example, if it's u<90, or u>90, then it's a one-tailed test. If it's u [not =] 90, then it's a two-tailed test. Clearly, this first question asks about a one-tailed test. Given the null and alternative hypothesis, and the significance level (in your case, 0.05), we need to find the rejection region. This will be an interval such that if the observed sample mean lies within it, then it will be "safe" (a 95% confidence level in this case) to conclude that the null hypothesis is false and reject in favor of the alternative hypothesis. In this case, the null is u=90 the alternative hypothesis is u>90. Therefore, we would reject the null hypothesis if the actual sample mean were "much greater" than 90, and assume that the null hypothesis is true. Specifically, the rejection region, will "start" in a number (let's call it 'a') such that if the actual mean were really 90, then the probability of observing a sample mean greater than 'a' would be 0.05. So let's write the math for this. Let's call Xbar to the observed sample mean. We want find 'a' such that: Prob (Xbar > a) = 0.05 We don't know the distribution of Xbar in order to solve this. However, we can make use of the Central Limit Theorem. This theorem states that, for a large sample size (usually more than 30 observations), (Xbar - m)/ (s/sqrt(N)) follows approximately a standard normal distribution; where m is the population mean, s is the population standard deviation and sqrt(N) is the square root of the sample size. In your case, we know that s = 22.5494... and sqrt(N) = 10; so s/sqrt(N) = 2.25494... Now, using this theorem, let's "convert" the Xbar in our equation into a standard normal distribution. Recall we're using that the population mean is actually equal to 90, so m=90: Prob (Xbar > a) = 0.05 Prob (Xbar - 90 > a - 90) = 0.05 Prob ( (Xbar - 90)/2.25494 > (a - 90)/2.25494 ) = 0.05 Given the Central Limit Theorem, we now know that the left-hand side of the inequality follows a standard normal distribution. Let's rename it as Z: Prob ( Z > (a - 90)/2.25494 ) = 0.05 We can now solve this equation. Looking up in a standard normal distirbution table (which can be found in most Statistics books), we get that: Prob(Z > 1.64) = 0.05 [Please do let me know through a Clarification Request if you don't know how to use these tables] This 1.64 is the critical z-score. If the test statistic (which we'll see how to compute in a few moments) is greater than this z-score, then we'll reject the null hypothesis. If it's smaller, then we won't reject the null hypothesis. This critical z-score can also be used in the following way. We found that: Prob ( Z > (a - 90)/2.25494 ) = 0.05 Prob(Z > 1.64) = 0.05 Therefore, (a - 90)/2.25494 = 1.64 a = 1.64*2.25494 + 90 a = 93.69... This implies that if the observed sample mean is greater than 93.69, then we'll reject the null hypothesis. In this case, we're told that that the sample mean is 94.4. We thus REJECT THE NULL HYPOTHESIS. We can also use test statistic in order to arrive to the same conclusion. It's calculated as: (Observed Sample Mean - Mean from Null hypothesis)/ (s/sqrt(N)) = (94.4 - 90) / 2.25494 = 1.9512 As you can see the test statistic is greater than the critical z-score, therefore, we reject the null hypothesis. Question 2 The logic in this test is very similar to the previous one, with the difference that we now need to use a two-tailed test. As such, the rejection region will now be "spli in two": we'll reject the null hypothesis if the observed sample mean is "too large" but also if it's "too small". We now want would like two values 'a' and 'b' such that: Prob(Xbar > a) = 0.025 Prob(Xbar < b) = 0.025 The 0.025 comes from the fact, in two-tailed tests, we need to divide alpha (0.05) by 2, because there are two "separate" rejection regions. By doing this, the sum of the probabilities in the two rejection regions will be 0.05, which is the desired significance level. We now proceed exactly as we did in the previous question. We'll start using Prob(Xbar > a) = 0.025 We "convert" it to standard normal just as we did before: Prob( (Xbar - 90)/2.25494 > (a - 90)/2.25494 ) = 0.025 Prob( Z > (a - 90)/2.25494 ) = 0.025 Looking up in a table, we get that: Prob( Z > 1.96 ) = 0.025 Therefore, the POSITIVE CRITICAL Z-SCORE IS 1.96. Alternatively, we can say that the "upper" rejection region are the values for the sample mean greater than 90 + 1.96*2.25494 = 94.42. Since the sample mean is smaller than this level (94.4) we still can't reject the null hypothesis - we'll arrive to the same conclusion later with the test statistic. Let's now calculate the other rejection region. We had: Prob(Xbar < b) = 0.025 Doing the same calculations, we get: Prob( Z < (b - 90)/2.25494 ) = 0.025 We don't need to use a table for this. Since the standard normal distribution is symmetric around zero, it has the following property: Prob ( Z > x) = Prob( Z < -x) Therefore, since the probability is 0.025 in both cases, then we conclude that the critical z-score from this equation is the opposite of the one we found earlier. Specifically, the NEGATIVE CRITICAL Z-SCORE IS -1.96. We can also compute the lower rejection region in terms of the sample mean just as we did before: (b - 90)/2.25494 = -1.96 b = 90 - 1.96*2.25494 = 85.58 This means that if the sample mean were lower than 85.58, we would reject the null hypothesis. Since it's neither lower than 85.58 nor greater than 94.42 (it's 94.4), then we conclude that we CANNOT REJECT THE NULL HYPOTHESIS. Let's find the same solution with the test statistic. This statistic is calculated in excatly the same way as before: (Observed Sample Mean - Mean from Null hypothesis)/ (s/sqrt(N)) = 1.958 Since the test statistic is neither greater than the positive critical z-score, nor lower than the negative critical z-score, then we conclude that we CANNOT REJECT THE NULL HYPOTHESIS. I hope this helps! If you have any doubt regarding my answer, please don't hesitate to request clarification before rating it. Otherwise, I await your rating and final comments. Best wishes! elmarto```