Question

With the following dataset, calculate the mean. Write out the null hypothesis using this mean, as well as the alternate hypothesis. Test the hypothesis, and explain your rejection or failure to reject the null. Null hypothesis: Does the average American spend more than $277.50 a week?

Ho: mu> 277.5

140 500 240 300 90 470 720 420 210 80 60 100

Answer 1

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho :u=277.5 against alternative hypothesis H1 :p> 277.5 Let i denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, T= 277.500, s2 = 43729.545455,8 = 209.12, n = 12 The test statistic can be written as: V/m (3 – 277.5) te s 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 > t0.05,n-1 = 1.795885 0.4- 0.3- 0.2- 0.1- 0.0- -2 Now, Value of the test statistic: The value of the test statistic tobs 12 (277.5000 - 277.5) 209.1161 = 0.000000 0.00 associated degrees of freedom = 11 The critical value = 1.795885 1.80 P-value = P(tu > 0.000000) = 1 - P(t1 < 0.000000) = 1-0.500000 =0.500000 0.5000 Conclusion : Since p-value > 0.05 and tobs * tcritical = 1.795885, so we fail to reject Ho at 0.05 level of significance Hence, we can conclude that population mean is not significantly greater than 277.5

There is not sufficient evidence to support the claim that American spend more than $277.50 a week.