Hypothesis: Standard Deviation and Critical Value

profit by entering that market. If your bikes are not of higher quality than the average bike already in the market, you will make a loss by entering the market for bikes. You test, at the 5% level of significance, the hypothesis that the quality of your bikes exceeds the average quality of existing bikes in the market. Given this information, answer the following questions. (a) What would a Type I Error be in the context of this case? Explain what the consequences of a Type I Error would be. (2 marks) (b) What would a Type II Error be in the context of this case? Explain what the consequences of a Type II Error would be. (2 marks) (c) After conducting the test, you conclude that there is sufficient evidence in the data you observe to support the alternative hypothesis that

A null hypothesis is that the average pulse rate of adults is 70. For a

With the P-value of 0.0033 being less than the alpha at 0.05 we will reject the null

The p-value for the hypothesis testing was lower than 0.05, thus, the null hypothesis is rejected and accepting the alternative hypothesis.

The state with highest population has size of 37,252,000. It is assumed that the mean of the sample population is less than 38,000,000. The hypothesis test was conducted to verify that the mean of the sample population is sample less than 38,000,000. Since it doesn’t contain the equal sign, it would be the alternate hypothesis. Let us assume that µ0 = 38,000,000 is the mean of the sample.

9. Suppose that you were asked to test H0: μ = 10 versus Ha: μ > 10 at the [pic] = 0.05 significance level and with a sample of size n = 10. Furthermore, suppose that you observed values of the sample mean and sample standard deviation and concluded that H0 be rejected. Is it true that you might fail to reject H0 if you were to observe the same values of the sample mean and standard deviation from a sample with n > 10? Why

Based on the presented problem above, the hypothesis will be formulated and tested at 0.05 level of significance.

After forming the hypothesis, the errors that can occur with respect to the hypothesis are:

variances are equal. The F-value is 1.93 and the critical F value is 1.61. Because

To verify the claim that the bottles contain less than sixteen ounces of soda a hypothesis test will be performed. To find the null hypothesis that represents the claim to be true or to be used as a basis of argument until it is proven. The null hypothesis is H0 greater than or equal to sixteen ounces. The alternative hypothesis is H1 less than sixteen ounces. The type of test used is one tailed testing. If using a significance level of .05, a one-tailed test allots the entire alpha to testing the statistical significance in the one direction of interest.  This means that .05 is in one tail of the distribution test statistic. When using a one-tailed test, it is testing for the possibility of the relationship in one direction and completely disregarding the possibility of a relationship in the other direction ("UCLA: Statistical Consulting Group," 2007). The value of the test is calculated by the mean 1 minus mean 2. Therefore, 14.87 – 16 equals -1.13 divided by standard deviation 0.55 divided by square root of random sample 30 equals .375. The P value is calculated by using the formula P (Z < equal to .375). The P value is 0.646. The P value is greater than the significance level of test which is .05. The answer is the mean of 0.5 < 0.646. The conclusion of the test for the null hypothesis is not rejected. This suggests the alternative hypothesis must be true that the soda in the bottles is less than sixteen ounces. The type of error

The distribution of the test statistic under the null-hypothesis is derived from the assumptions identified previously. Common test statistics may follow the following distributions: Normal, Student T, and Chi-Square. This distribution separates the possible values of the estimator into two categories: values for which the null-hypothesis is accepted or rejected. The region for which we accept the null-hypothesis is called the critical region and the area underneath the curve that corresponds to the critical region is known as the level of confidence. Hence, we can develop a confidence interval for which we can see the lowest and highest point of the critical region. Any observed sample mean that lies outside of this confidence interval (outside the critical region) would cause us to reject the null-hypothesis in favor of the alternative hypothesis. The area of the rejection region is known as the level of significance and represents type I error (alpha) corresponding to the probability that a true null-hypothesis is rejected (as opposed to type II error- beta; the probability of accepting a false null-hypothesis). Essentially, hypothesis testing calls for comparing a test statistic to the critical value of the test statistic. If this test statistic is greater than the critical value of the test statistic, we will reject the null hypothesis in favor of the alternative hypothesis. If

they imply that we cannot reject the null hypothesis of equal means. Also notice the p_value of

The critical value is v = 44 + 56 – 2 = 98. Using the student’s t distribution, at a = 0.05, t = 1.985. The decision rule reject Ho if t 1.985 otherwise do not reject Ho.

The test statistic is -34.10940 which is more negative than the critical value and hence

There are four ideas to keep in mind when creating the null and alternative hypotheses :