Question

9. Medicare would like to test the hypothesis that the average monthly rate for a one- bedroom assisted-living facility is equal to $3,300. A random sample of 12 assisted-living facilities had an average rate of $3,690 per month. The standard deviation for this sample was $530. Medicare would like to set a = 0.05. +$3,300 and it is a Note: Two hypotheses for the test are H :u=$3,300 vs H : two-tailed test. Calculate the test statistics • find the critical values and mark on the corresponding curve. Mark the rejection regions. • Would you reject the null hypothesis? Give the reason for your decision.

Answer 1

Let u denotes the population mean. Hypotheses : To test null hypothesis H :p =3300 against alternative hypothesis H1 :#3300 Let i denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, ž= 3690.000, s? = 280900.0000, s = 530.000, n= 12, standard error of mean = S 530.00 = VTV12 - = 152.9978 The test statistic can be written as: na – 3300) which under H, follows a t distribution with n - 1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs | > to.025,n-1 = 2.200985 density -4 -2 N- Now, Value of the test statistic: The value of the test statistic tobs = 12 (3690.0000 - 3300) -= 2.549056 530.0000 2.55 associated degrees of freedom = 11 The critical value = Ft 0.05 = Eto.025.11 = +2.200985 +2.2010 P-value = P(tn-11 > tobs) = 2* P(t11 > 2.549056) = 2 + (1 -0.986480) = 0.027041 0.0270 Conclusion : Since p-value < 0.05 and tobs > tcritical = 2.200985, so we reject H, at 0.05 level of significance Hence, we can conclude that the the population mean is significantly different from 3300.