The number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 2.05 entrees per order. On a particular Saturday afternoon, a random sample of 50 Noodles orders had a mean number of entrees equal to 2.4 with a standard deviation equal to 0.82. At the 5 percent level of significance, does this sample show that the average number of entrees per order was greater than expected?(a) Choose the correct null and alternative hypotheses.a. H0: μ ≥ 2.05 vs. H1: μ < 2.05b. H0: μ ≤ 2.05 vs. H1: μ > 2.05c. H0: μ = 2.05 vs. H1: μ ≠ 2.05

Answer: b. [tex]H_0:\mu=2.05[/tex] vs [tex]H_1: \mu>2.05[/tex]Step-by-step explanation:Let [tex]\mu[/tex] be the population mean.Given : The number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 2.05 entrees per order.[tex]\mu=2.05[/tex]Claim : The average number of entrees per order was greater than expected.i.e.[tex]\mu>2.05[/tex]Hence, the  correct null and alternative hypotheses for the given description:-[tex]H_0:\mu=2.05[/tex][tex]H_1: \mu>2.05[/tex]   [Alternative hypothesis shows significance difference.]Since , the alternative hypothesis is right-tailed , so the test is a right tailed test.Test statistic : [tex]z=\dfrac{2.4-2.05}{\dfrac{0.82}{\sqrt{50}}}\approx3.02[/tex]p-value : P(z>3.02)=0.0012639Since p-value (0.0012639) is less than the significance level (0.05) , so we reject the null hypothesis.Conclusion : The average number of entrees per order was greater than expected