Hypothesis Test for Lightning Deaths Refer to the sample data ...

Question

Hypothesis Test for Lightning Deaths Refer to the sample data given in Cumulative Review Exercise 1 and consider those data to be a random sample of annual lightning deaths from recent years. Use those data with a 0.01 significance level to test the claim that the mean number of annual lightning deaths is less than the mean of 72.6 deaths from the 1980s. If the mean is now lower than in the past, identify one of the several factors that could explain the decline.

Accepted Answer

Hello, everyone. Let's take a look at this question together. Consider the following data representing the numbers of annual deaths from heat waves in the United States over 20 consecutive years. 54 49 47 53 50 36 42 46 41 33 38 3129 35 21 30 3244, 17 and 22. Assume these data are a random sample of recent annual heat wave deaths at the 0.01 significance level, test the claim that the mean number of annual heat wave deaths is less than the mean of 80.2 deaths from the 1960s. If the mean is now lower, mention one factor that might explain the reduction. So, from the information provided in the question, our given data includes the annual deaths from heat waves over the last 20 years with the numbers recorded, a sample size represented by N of 20. A claimed past mean of mu sub-zero, which equals 80.2, a significance level of alpha equals 0.01, and the claim that the mean is now less than 80.2. Which this is a left tailed test. And so the first step in testing this claim is to calculate the sample statistics. Starting with the sum of all of the data where we add all of the recorded death numbers together, which gives us 750 as the sum of the data and using the sum we can calculate the sample mean which. Represented by X bar, which is equal to 750 divided by 20, which gives us 37.5 as our sample mean, and then we can calculate our sample standard deviation, which can be calculated using the formula for the sample standard deviation where we calculate the values for the formula into the following table. And then substituting the values into the sample standard deviation formula, we get our sample standard deviation of 10.763. And then we can define our hypotheses where our null hypothesis is that mu is equal to 80.2, whereas our alternative hypothesis is that mu is less than 80.2. So now we can calculate our test statistic. Using the following formula for the test statistic T, which is equal to X minus mu subzero divided by S divided by the square root of N. And substituting the values that we have calculated into the test statistic formula, we get a test statistic of approximately -17.72. And then we can determine the critical T value. Where our degrees of freedom is equal to N minus 1, so since N is 20, our degrees of freedom is 19. And for a left tailed T test at an alpha of 0.01, using a T distribution table, we get our T critical value of -2.539. And using our T critical value of -2.539, we have to make our decision. And since our T statistic of -18.42 is less than our critical T value of -2.539, we know that the test statistic is far into the rejection region. Therefore, we reject the null hypothesis. So in conclusion, at the 0.01 significance level, we reject the null hypothesis, as there is strong evidence that the mean number of annual heat wave deaths. is now less than the mean of 80.2 from the 1960s. And so answer choice B is the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Hypothesis Testing

Significance Level

Mean Comparison

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