A marketing expert for a pasta-making company believes that 50% of pasta lovers preferlasagna. If 13 out of 20 pasta lovers choose lasagna over other pastas, what can beconcluded about the expert's claim? Use a 0.01 level of significance.Click here to view the binomial probability sums table for n=17 and n=18.Click here to view the binomial probability sums table for n=19 and n=20.Let a success be a pasta lover that chooses lasagna over other pastas. Identify the nulland alternative hypotheses.OA. Ho: p > 0.5H₁: p=0.5D. Ho: p<0.5H₁: p=0.5B. Ho: p=0.5H₁: p0.5E. Ho: p=0.5H₁: p<0.5The test statistic is a binomial variable X with p(Type integers or decimals. Do not round.)and n =C. Ho: p= 0.5H₁:p> 0.5F.Ho: p0.5H₁: p=0.5 Past experience indicates that the time required for high school seniors to complete astandardized test is a normal random variable with a standard deviation of 7 minutes.Test the hypothesis that o = 7 against the alternative that o<7 if a random sample of thetest times of 18 high school seniors has a standard deviation s = 5.83. Use a 0.10 levelof significance.Click here to view page 1 of the Chi-Square Critical Values Table.Click here to view page 2 of the Chi-Square Critical Values Table.Identify the null and alternative hypotheses. Select the correct choice below and fill inthe answer boxes to complete your choice.(Type integers or decimals. Do not round.)A.C.Ho: 0²>H₁:1:0² =Ho: 0²=H₁:0²>B. Ho: 0²<0² =H₁:D. Ho: 0²:=H₁:0² <

"As per the Bartleby guidelines, we can solve only one question at a time.Therefore, we are solving only the first one for you."We have to test the claim that of the pasta lovers prefer lasagna over pasta at For testing the given claim, the sample information is given as:Number of pasta lovers taken in the sample: Out of this, number of persons who choose lasagna: The null hypothesis is the statement which is tested for the possible rejection and alternative hypothesis is the statement contradicting the null statement.Therefore, the null hypothesis is assumed as the population proportion is equals to 0.50, that can be stated as:which is tested against the alternative hypothesis that the population proportion is not equals to , that can be stated as:Option b is the correct set of hypotheses.As the alternative hypothesis contains not equals to sign, it is a two tailed test.Rest of the options are incorrect as they do not match correctly with the obtained outcomes for hypotheses.The test statistic X which denotes the number of persons who prefer lasagna over pasta follows binomial distribution with parameters as .The sample proportion is obtained as:Therefore, the value of test statistic is obtained as:The value of test statistic is obtained as .The two-tailed p-value for the test is obtained as:The p-value is obtained as .Note: The is obtained using the MS-excel formula "=NORMSDIST( )" and the required probability is obtained as:Decision Rule: If the obtained p-value is less than the level of significance, we will reject the null hypothesis.Otherwise, we fail to reject the null hypothesis.Therefore, as the obtained p-value is greater than the (0.01), we fail to reject the null hypothesis and we can not conclude that the proportion of people who prefer lasagna over pasta is different from 50%.Option b is the correct set of hypotheses, which are:The test statistic is 1.3416.The p-value is 0.1798.At , we fail to reject the null hypothesis and we can not conclude that the proportion of people who prefer lasagna over pasta is different from 50%.