Question

A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers. Assume the population standard deviation for those drinking regular coffee is 1.45 cups per day and 1.56 cups per day for those drinking decaffeinated coffee. A random sample of 47 regular-coffee drinkers showed a mean of 4.42 cups per day. A sample of 37 decaffeinated-coffee drinkers showed a mean of 4.93 cups per day. Use the 0.025 significance level. Is this a one-tailed or a two-tailed test? One-tailed test. Two-tailed test. State the decision rule. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) What is the p-value? What is your decision regarding H0? Do not reject H0. Reject H0.

Answer 1

Answer:

• Null and alternative hypothesis is $\small H_0: \mu_1\geq \mu_2\ \ \ \verses\ \ H_1: \mu_1< \mu_2$
• One tail test ( left tail )
• critical value = $\small Z_c=-1.96$
• decision criteria : if test statistic z < $\small Z_c=-1.96$ then we reject $\small H_o$ other wise we accept.
• test statistic = z= -1.53
• p-value= 0.0625
• Decision : Do not reject $\small H_o$

Answer: Given, claim: whether the mean daily consumption of pregular a coffee cookens is less than _that of decaffeinated coffeechinkers... let sample one for regular Coffee drinkers _ sample too for decaffeinated coffee t doinkers 61 = population standard deviation for - regular coffee drinkers 6 = 1045 cups 62 = population standard deviation for decaffeinated coffee cookers 62 = 1156 cups. X = 4.42 X 2 = 4.95 ni = 47__ D2 = 32 Q=0.025 from claim. Ho: el >ll2 1. Hi: el <42 Null bypothesis Alternative bypother's Here Alternative bypothesis is less than so test is one tailed test. left tailed test wherre, one failed to Now, here we use z-test because population standard deviation known.

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Decision - test statistic Z < -1.96 then reject Ho. (.' ZC = 20.025_2 -1.96_ from normaltable left tailed. test statistic I = Xi - X2 tortor ta na 4:42 - 4.93 1.452 1.562 47 37 =-1153 test Statistic = -1:53 p-value: _p-value = 0.0625_ From Normal table Z=-1.53 Decision : p_value = 0.0625 Y R=0.025 therefore Do not rejeet Null bypothesis.