Using the attached data, we like to test that mean price of neutral is lower tha

Question

Using the attached data, we like to test that mean price of neutral is lower than mean price of sad with 0.05 significance. Do manually a test using the assumption. neutral sad 1.00 2.50 3.00 4.00 1.00 1.50 4.00 1.00 1.50 3.50 . 0.50 0.50 3.00 1.50 2.00 0.00 3.50 2.50 0.00 1.00 2.00 0.00 0.00 0.00 1.00 2.00 0.00 0.00 (note: you need to set up, compute statistics, cv from a table, conclusion, implication and error statements for each test.) 1) we assume that variance of neutral = 0.6 and variance of sad = 1.3 2) We assume that variances are unknown but same 3) We assume that variances are unknown and not same 4) do #2 and #3 again using ttest in R studio. Just write down conclusion using p-value (note: in ttest, you need to put "mu = 0") 5) do shapiro test to see that data are from normal 6) write down implication of shapiro test for validity of t test 1.00 0.00 0.50

Explanation / Answer

Question 1

Solution:

Here, we have to use two sample z test for population mean.

H0: µ1 = µ2 versus Ha: µ1 < µ2

This is a one tailed – lower tailed test.

We are given = 0.05

Test statistic is given as below:

Z = (X1bar – X2bar) / sqrt[(12 / n1)+(22 / n2)]

From given data, we have

X1bar = 0.571429

X2bar = 2.117647

1 = 0.6

2 = 1.3

n1 = 14

n2 = 17

Z = (0.571429 - 2.117647) / sqrt[(0.6^2/14)+(1.3^2/17)]

Z = -4.3712

P-value = 0.00

P-value < = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that mean price of neutral is lower than mean price of sad.

Question 2

Here, we have to use two sample t test assuming equal variances.

H0: µ1 = µ2 versus Ha: µ1 < µ2

We are given = 0.05

Test statistic is given as below:

t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]

Where Sp2 is pooled variance

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

X1bar = 0.571429

X2bar = 2.117647

n1 = 14

n2 = 17

S1 = 0.730046

S2 = 1.244104

df = 14 + 17 – 2 = 29

Sp2 = [(14 – 1)*0.73^2 + (17 – 1)*1.24^2]/(14 + 17 – 2)

Sp2 = 1.0929

(X1bar – X2bar) = 0.571429 - 2.117647 = -1.5462

t = -1.5462 / sqrt(1.0929*((1/14)+(1/17))

t = -4.0982

P-value = 0.0002

P-value < = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that mean price of neutral is lower than mean price of sad.

Question 3

Here, we have to use two sample t test assuming unequal variances.

H0: µ1 = µ2 versus Ha: µ1 < µ2

We are given = 0.05

Test statistic is given as below:

t = (X1bar – X2bar) / sqrt[(S12 / n1)+(S22 / n2)]

X1bar = 0.571429

X2bar = 2.117647

n1 = 14

n2 = 17

S1 = 0.730046

S2 = 1.244104

df = 26 (by using excel formula)

(X1bar – X2bar) = 0.571429 - 2.117647 = -1.5462

sqrt[(S12 / n1)+(S22 / n2)] = sqrt[(0.73^2/14)+(1.24^2/17)] = 0.3593

t = -1.5462 / 0.3593

t = -4.3031

P-value = 0.0001

P-value < = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that mean price of neutral is lower than mean price of sad.

Question 4

Required R commands are given as below:

> Neutral = c(0,1,2,0,0,0,1,2,0,0,0.5,1,0,0.5)

> Sad = c(1,2.5,3,4,1,1.5,4,1,1.5,3.5,0.5,3,1.5,2,0,3.5,2.5)

> t.test(Neutral, Sad)

        Welch Two Sample t-test

data: Neutral and Sad

t = -4.3031, df = 26.48, p-value = 0.0002046

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-2.2841749 -0.8082621

sample estimates:

mean of x mean of y

0.5714286 2.1176471

P-value < = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that mean price of neutral is lower than mean price of sad.

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