use calculator method ti 83 or 84 please 2.36 Downloading music. A husband and wife, Stan andLucretia, share a digital music player that has a featurethat randomly selects which song to play. A total of 2643songs have been loaded into the player, some by Stan andthe rest by Lucretia. They are interested in determiningwhether they have loaded different proportions of songsinto the player. Suppose that when the player was in therandom-selection mode, 27 of the first 40 songs selectedstep22.394stepwere songs loaded by Lucretia. Let p denote thetion of songs that were loaded by Lucretia.propor-(a) State the null and alternative hypotheses to betested. How strong is the evidence that Stan andLucretia have loaded different proportions of songsinto the player? Make sure to check the conditionsfor the use of this test.(b) Are the conditions for the use of the large-sampleconfidence interval met? If so, estimate with 95%confidence the proportion of songs that were loadedby Lucretia.

Given: The total number of songs are 2643. Let p represents the population proportion of songs. Let p^ represent the estimated proportion of songs. Since the hypothetical value (p0) is not given here, we will assume that it is 0.5. Lucretia loaded 27 of the first 40 songs; that is p^=2740 The conditions for the use of this test are: 1) The sample is selected by simple random sampling from a population with population proportion. 2) The data is categorized into two mutually exclusive classes. Here, the population proportion of songs is randomly selected with given population proportion. So, this condition is valid . Also, the intersection of selecting different songs and selecting similar proportion of songs is mutually exclusive. Hence, the conditions are satisfied to use this test. (a)The hypothesis that will be formed here is: Null hypothesis (H0) : p=0.5 ( There is no difference in proportions of songs) Alternate hypothesis( H1): p≠0.5 ( There is a significant difference in proportions of songs) The value of test statistics is given as: Z= p^-p0p^q^n =2740-0.50.5×0.540 =2.215 The calculated value is 2.215. Assuming the level of significance as 0.05. The tabulated value of the test statistics of two tailed test at 0.05 level of significance is given as 1.96. Hence, The calculated value is less than the tabulated value . Therefore, the null hypothesis is rejected. As a result there is a significant difference in population proportions of the songs. (b) The confidence interval will be given as: C.I=p^±zα2·p^q^n , =2740+ 1.96 2740×134040 =0.82015and C.I=p^±zα2·p^q^n , =2740- 1.96 2740×134040 =0.5299 Hence, the confidence interval 95% level of significance is given by (0.5299, 0.82015)