Choose for a model to follow that shows course expectations:Hypothesis Testing Example for a Single Proportion R.Andrusiak.pdf. (Attach)
Hypothesis Testing Example for a Single Proportion (R.Andrusink; inspired by student provided content) Research studies have shown that Americans prefer cake and pie equally. Shari decides to test these claims. When deciding on her hypotheses Shari believes that Americans tend to prefer pie over cake (hence the saying \"As American as Apple Pie"): however. she realizes that she does not have any support for that belief and since she does not want to overlook one preference over the other. she decides to conduct a two-sided test. She makes the following hypotheses. Hg: p = 0.50. where p represents the proportion of all Americans that prefer pie over cake. HA: p a: 0.50. where p represents the proportion of all Americans that prefer pie over cake tie, the proportion of Americans that prefer pie over cake is different from SWooould be less than or greater than 50%}. Shari conducts a simple random sample of 563Amerielm {don't ask her how as it would be really difficult to do in real life), and nds that that 312 of than prefer pie over cake. Does Shari have support for or against her null hypothesis? First check conditions: Independence: The independence condition is satised as this was a simple random sample. Successi'ailure Condition: (Important note: Notice that the null hypothesis value, pa = 0.51]. is used in this calculationwe are hypothesizing tilt the distn'bution of the sample proportion should center at 950; thus, we build the distributitut based upon this assuraption and then see where the sample proporl. :3. falls on that distrillOn. trip = 563(0.50) 3: 10 nfl Po) = 563(1 use) 2 10 With both conditions satised, we can move forward. If the null hypothesis were true, the sampling distribution of the mic proportion should follow a normal distribution centered at 0.50 with standard mowers, = W?\" = I? e on21. This leads to this distribution: 20 0.4 0.42 0.44 0.56 0.58 0.6 (0.5,0 P 0.542,0 (0.458,0 (0.479, 0.P (0:521, 015 We want to see how likely it is, if the null hypothesis were true, to come up with a value of p = 563 312 ~ 0.554 or greater or 0.446 or less, by random chance alone. [Note: since this is a two-sided test (based upon the alternative hypothesis being a not-equal situation), we need to consider values just as extreme on the other side of the distribution. Thus, I look at 0.554 at being 5.4% above the hypothesized center of 50%, so I went 5.4% below the hypothesized center of 50% to get 44.6%.] To determine this likelihood, we find the =-score-how many standard errors away from the mean is my point-estimate. point estimate - null value 0.554 - 0.50 z = ! 2.56 standard error 0.50(1 - 0.50) 563 This graph shows where the point estimate lands and shows the area at 0.554 and above (shaded).0.42 0 56 0.58 0.6 (0.479.0 (0,542.0) to.3210 We now need to find the probability that = is at least 2.56. You can use the lookup table in the book, Desmos, or one of the other tools I pointed you to. Here is the probability based upon using Desmos. normaldist( a,b) X Mean, Standard Deviation Find Cumulative Probability (CDF) Min: 0.554 Max: 0 = 0.0050639952747 Since this is a two-sided test, to determine the p-value, we need to consider the same area on the other side of the distribution (0.446 or less for our point-estimate). However, due to the symmetry in the distribution, we will get the same value as above. Therefore, the p-value is about 2(0.00506) = 0.01012. Note, this is slightly bigger than 1%, but less than 5%. If we are using the significance level of 1%, our result would not be statistically significant (i.e., do not reject the null hypothesis as the p-value is not less than 19%). If we were using the 5% significance level, the result is statistically significant (as it is less than 5%) and we would havesupport for the alternative hypothesis. We set the significance level before collecting data (not after). Let's assume we set it at the 5% level. In this case, it means that we have support for the alternative hypothesis, that the proportion of Americans that prefer cake over pie is different from 50% (seems as if Americans are preferring pie over cake, but we didn't run a one-sided test). Here is the technically statement of the interpretation which always follows this format: If the null hypothesis is true (state it in context here), then the probability of finding (state what you found) by random chance alone is about (insert p-value here). Therefore, we should (state if you reject or do not reject the null hypothesis and what this means in context). Here it is: If it is true that Americans prefer pie and cake equally, the probability of finding 55.4% or more or 44.6% or less Americans who prefer pie over cake, in a random sample of 563 Americans, by chance alone is about 1.012%. This is significant at the 5% level. Therefore, we reject our null hypothesis and have support for the alternative hypothesis that the proportion of Americans who prefer pie over cake is different from 50%
