observed expected
Brand 1: Bridgestone |
131 |
200 |
Brand 2: Callaway |
273 |
200 |
Brand 3: Nike |
119 |
200 |
Brand 4: Srixon |
301 |
200 |
Brand 5: Titleist |
176 |
200 |
Brand 6: Wilson |
200 |
200 |
total |
1200 |
1200 |
137.14 |
chi-square |
|
5 |
df |
|
7.25E-28 |
p-value |
(a)
Ho: The proportion of boxes sold are equal for all 6 brands. i.e. 1/6
Pbridgestone = 1/6, Pcallaway = 1/6, Pnike = 1/6, Psrixon = 1/6, Ptitlesit = 1/6, Pwilson = 1/6
Ha: Some of the population proportions differ from the values stated in the null hypothesis
This corresponds to a Chi-Square test for Goodness of Fit.
(b)
= 137.14
critical = 11.070
As (137.14) is greater than critical, we reject the Null hypothesis.
Also as the p value ( 7.25E-28) is less than (0.05) we reject the Null hypothesis.
Hence we have sufficient evidence to believe some of the population proportions differ from the values stated in the null hypothesis.
Sales (measured by number of boxes sold with 12 balls in each box) of six different...
You are conducting a multinomial Goodness of Fit hypothesis test for the claim that the 4 categories occur with the following frequencies: Ho : pA=0.3; pB=0.15; pC=0.15; pD=0.4 Complete the table. Report all answers accurate to three decimal places. Category Observed Frequency Expected Frequency A 29 B 41 C 12 D 53 What is the chi-square test-statistic for this data? χ2= What is the P-Value? P-Value = For significance level alpha 0.05, what would be the conclusion of this hypothesis...
please provide a detailed answer, thank you! B1 pts Over You are conducting a Goodness of Fit hypothesis test for the claim that the population data values are not distributed as follows: PA = 0.4; P = 0.25; Pc = 0.1; P = 0.25 Give all answers as decimals rounded to 3 places after the decimal point, if necessary. (a) What are the hypotheses for this test? OH:PA = 0.4, PB = 0.25, Pc = 0.1, PD = 0.25 H:...
The frequency distribution shows the results of 200 test scores. Are the test scores normally distributed? PART B. Determine the critical value and the rejected region PART C. Calculate the test statistic PART D. Decide whether to reject or fail to reject the null hypothesis The frequency distribution shows the results of 200 test scores. Are the test scores normally distributed? Use α= 0.01. Complete parts (a) through (d) Class boundaries Frequency, f 49.5-58.5 20 58.5-67.5 62 67.5-76.5 79 76.5-85.5...
The frequency distribution shows the results of 200 test scores. Are the test scores normally distributed? Use a =0.01. Complete parts (a) through (e). Class boundaries 49.5-58.5 58.5-67.5 Frequency, f 19 62 D 67.5-76.5 81 76.5-85.5 33 85.5-94.5 5 Using a chi-square goodness-of-fit test, you can decide, with some degree of certainty, whether a variable is normally distributed. In all chi-square tests for normality, the null and alternative hypotheses are as follows. Ho: The test scores have a normal distribution....
The frequency distribution shows the results of 200 test scores Are the test scores nomally distributed? Use α-0.10 Complete parts is) through (e) Class boundaries Frequency, 49.5-585 2D 58.5-67.5 51 67.5-76.5 B0 76.5-85.5 35 95.5-94.5 Using a chi-square goodness-of-fit test, you can decide, with some degree of certainty whether a variable is normally distributed. In all chi-square tests for nomality the null and alternative hypotheses are as follows Ho The test scores have a nomal distribution Ha The test scores...