Question

A study investigated whether price affects people's judgment. Twenty people each tasted six cabernet sauvignon wines and rated how they liked them on a scale of 1 to 6. Prior to tasting each wine, participants were told the price of the wine. Of the six wines tasted, two were actually the same wine, but for one tasting the participant was told that the wine cost $10 per bottle and for the other tasting the participant was told that the wine cost $90 per bottle. The participants were randomly assigned either to taste the $90 wine first and the $10 wine second, or the $10 wine first and the $90 wine second. Differences were calculated by subtracting the rating for the tasting in which the participant thought the wine cost $10 from the rating for the tasting in which the participant thought the wine cost $90.Difference ($90 − $10):3 , 4 ,2 ,2 , 1, 0, 0, 3, 0, 2, 1, 3, 3, 1, 4, 1, 2, 2, 1, −1Carry out a hypothesis test to determine if the mean rating assigned to the wine when the cost is described as $90 is greater than the mean rating assigned to the wine when the cost is described as $10. Use α = 0.01. (Use a statistical computer package to calculate the P-value. Use μ$90 − μ$10. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places)

Answer 1

To determine if the mean rating assigned to the wine when the cost is described as $90 is greater than the mean rating assigned to the wine when the cost is described as $10, a hypothesis test is conducted.

Step-by-step explanation:

To determine whether the mean rating assigned to the wine when the cost is described as $90 is greater than the mean rating assigned to the wine when the cost is described as $10, a hypothesis test is conducted. The null hypothesis assumes that the mean rating difference is not greater than zero, while the alternative hypothesis assumes that the mean rating difference is greater than zero.

A one-sample t-test is used, and the test statistic is calculated by dividing the mean difference of the ratings by the standard error. The degrees of freedom are determined by the sample size minus 1.

The resulting test statistic is compared to the critical value from the t-distribution with the given alpha level. If the test statistic is greater than the critical value, the null hypothesis is rejected, indicating that there is sufficient evidence to conclude that the mean rating assigned to the $90 wine is greater than the mean rating assigned to the $10 wine.

Answer 2

t=5.5080( to 3 d.p)

Step-by-step explanation:

From the data given,

n =20

Deviation= 34/20= 1.7

Standard deviation (sd)= 1.3803(√Deviation)

Standard Error = sd/√n

= 1.3803/V20 = 0.3086

Test statistic is:

t = deviation /SE

= 1.7/0.3086 = 5.5080

ndf = 20 - 1 = 19

alpha = 0.01

One Tailed - Right Side Test

From Table, critical value of t =2.5395

Since the calculated value of t = 5.5080 is greater than critical value of t = 2.5395, the difference is significant. Reject null hypothesis.

t score = 5.5080

ndf = 19

One Tail - Right side Test

By Technology, p - value = 0.000

Since p - value is less than alpha , reject null hypothesis.

Conclusion:

From the result obtained it can be concluded that ,the data support the claim that the mean rating assigned to the wine when the cost is described as $90 is greater than the mean rating assigned to the wine when the cost is described as $10.