STATISTIC

Statistics Test 2



Statistics Test 2

Q1 An experiment was conducted to explore the relationship between the height of an individual (X) and how much they weigh (Y). Sample data on 15 observations gave the following results for the linear equation  Y=-57.74+2.556X. The standard error of the slope coefficient is 0.98.  Calculate the 99% confidence interval for the slope coefficient:  2.556±..........(3dp).

Degrees of freedom = n-2

Where n = 15

Therefore, degrees of freedom = 15-2

degrees of freedom = 13

At 99% confidence interval, alpha level is 1%

Critical Probability = 1 - alpha/2

Critical Probability = 1 - 0.01/2

Critical Probability = 0.995

At Critical Probability = 0.995 and degres of freedom = 13, t-value has been taken from t-distribution table.

T-score = 3.012

Margin of Error (ME): ME = critical value * standard error = 3.012 x 0.98

Margin of Error (ME) = 2.952

99% confidence interval calculation

2.556 - 2.952 < Slope of Coefficient X < 2.556 + 2.952

-0.396 < Slope of Coefficient X < 5.508

We can be 99% confident that the slope of coefficient is in the range 2.556 ± 2.952.

Q2 A farmer conducted an experiment over 12 years to see if different types of fertilizer affected the yield of the field. He applied different fertilizer to each of the 3 fields for the entire time period and recorded the output each year. The sample gave the following information: the SSA, i.e. the variability attributed to the fertilizer was 189; the SSW, i.e the random variability of each of the fields output, was 313; the TSS, i.e. the total variability of the data, was 189+313. What is the value of the F statistic that would be used to test equality of the mean output? (3dp)

MSA = SSA / (n-1)

MSA = 189 / (3-1)

MSA = 189 / 2

MSA = 94.500

MSE = SSW / (n-k)

MSE = 313 / (36-3)

MSE = 313 / 33

MSE = 9.485

F- Statistics = MSA / MSE

F- Statistics = 94.500 / 9.485

F- Statistics = 9.963

Variation

Sum of Squares

df

Mean Square

F-Statistic

SSA

189

2

94.500

9.963

SSW

313

33

9.485

 

SST

502

35

 

 

Q3 If the F statistic calculated for testing the equality of 4 groups with a total sample size across the four groups of 34 is 2.90, would you reject the null hypothesis of equal means at the 5% level? Yes or No

From the F-distribution table, critical value of F-statistics is extracted as follows:

F0.05 (3, 30) = 2.9223

The test statistic has the F value of 2.90. From the F-distribution table, F-value is 2.9223. Because the test value of F-statistics is less than the critical value F-Value (2.90 < 2.9223), therefore, we accept the null hypothesis of equal means at 5% level. Therefore, we will not reject the null hypothesis.

Answer = No

Q4  One-way analysis of variance (ANOVA) is done to compare exam marks across three different modules in the Business programme.  What is the correct statement of the null hypothesis?

Null Hypothesis Correct Statement:

Mean exam marks across three different business modules in the Business programme is equal.

Numeric form of null hypothesis

H0: µ1 = µ2 = µ3

Abstract Form:

H0: µ1 (Mean exam marks in module 1) = µ2 (Mean exam marks in module 2) = µ3 (Mean exam marks in ...