Forecast and Actual Temperatures Listed below are actual tempe...

Question

Forecast and Actual Temperatures Listed below are actual temperatures (°F) along with the temperatures that were forecast five days earlier (data collected by the author). Use a 0.05 significance level to test the claim that differences between actual temperatures and temperatures forecast five days earlier are from a population with a mean of 0°F.

Accepted Answer

A nutritionist compares the actual calorie content of 8 meals with calories listed on the restaurant menu. The differences, actual minus listed in calories are 15, -10, 20, -5, 10, -15, 5, and -20. At the 0.05 significance level, Tesla claimed that the mean difference between the actual and listed calories is zero. We're also given 4 possible answers, whether you reject our hypothesis or fail to reject our hypothesis and determine if there's significant evidence. Now, to solve this, let's first set up our hypotheses. We have our mean is equals to 0. Or in other words, Mean difference Is 0. And for our alternative. We have at the mean is not equal to 0 or The main difference Is not 0. Now, we first need the mean and standard deviation to solve this. This table on the left here will help us find our standard deviation, but first, let's find our mean. I mean Which we will call X bar is given by the sum of all of our values, 15 plus negative 10. + 20 + 5. Plus 10 Plus -15. Plus 5 plus native 20. All divided by the sample size of 8 meals. We can simplify this to get 0 divided by 8 or just 0. Now our standard deviation We can first find by getting our variants. We want our squared deviations, which this table here gives us all of them. We'll go ahead and add all of these square deviations together, 225 plus 100, plus 400, and so on, until the last one of 400. This is then divided by our sample size minus 1 or 8 minus 1. Go simplify this to get 1500 divided by 7. Serin deviation then is just a square root of 1500 divided by 7. Or approximately 14.64. We now have mean inser deviation. Now that we have both of these, we can find our test statistic. We will use a T test for this one. So we have T equals. I mean minus the population mean. Divided by standard deviation, divided by the square root of sample sizes. In this case, our X bar is 0, and our hypothesized mean is also 0. So we have 0 minus 0 on the top. And the denominator of a fraction. We have 14.64, divided by the square root of 8. We get 0 divided by 5.18, which is just 0, meaning our test statistic is 0. We now need our critical T. Our degrees of freedom will be 8 minus 1 or 7. And we're at the 0.005 significance value. 0.05 significance. Now with a two-tailed test, T critical then, based on these parameters we can find from a table, but this table gives us plus or minus 2.365. Which means to reject the hypothesis, our T value should be outside of our interval, negative 2.365 to positive 2.365. However, our T is 0 and it's in between these two values. This means we fail. To reject Our no hypothesis. This means then There is not. Enough evidence. At the 0.05 level. To conclude A mean difference exists. Based on everything we said. We cannot conclude that there is a difference. That means the answer to this problem, if we look at our possible answers. I answer B. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Key Concepts

Hypothesis Testing

Significance Level

Mean Difference

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