Question

demographic markets – the Urban and Rural markets. As such, in each market, you run a...

demographic markets – the Urban and Rural markets. As such, in each market, you run a short survey that gauges customers demand for your product and assigns them to one of three categories – (i) High (ii) Medium or (iii) Low.

You survey 120 people in the “Urban” market and find that their demand falls into the following “buckets”

High 42

Medium 36

Low 42

You survey 80 people in the “Rural” market time. Their demand for the product its reflected below.

High 38

Medium 24

Low 18

The Null hypothesis that you are asked to test is that "the demand for the product is INDEPENDENT of the whether the market is Urban or Rural".

(A) Under this null hypothesis, what is the EXPECTED NUMBER OF PEOPLE THAT WILL ANSWER "HIGH" in the URBAN Market?

(B) Under this null hypothesis, what is the EXPECTED NUMBER OF PEOPLE THAT WILL ANSWER "LOW" in the RURAL Market?

(C) CALCULATE THE TEST STATISTIC FOR THE NULL HYPOTHESIS ABOVE. Enter that answer below as a number with two significant decimal places (such as 8.94 or 2.34 or 213.45)

(D) Using the critical values for the 5% and 1% levels of the Chi-Square Distribution from your text, which of the following statements is true?

We can reject the null hypothesis at the 1% level.

We can reject the null hypothesis at the 5% level.

We cannot reject the null hypothesis at the 5% or 1% level

We can reject the null hypothesis at both the 5% and the 1% levels.

Homework Answers

Answer #1
Column and Row Totals
Urban Rural Row Totals
High 42 38 80
Medium 36 24 60
Low 42 18 60
Column Totals 120 80 200  (Grand Total)
Results
Urban Rural Row Totals
High 42  (48.00)  [0.75] 38  (32.00)  [1.12] 80
Medium 36  (36.00)  [0.00] 24  (24.00)  [0.00] 60
Low 42  (36.00)  [1.00] 18  (24.00)  [1.50] 60
Column Totals 120 80 200  (Grand Total)
Numbers inside the bracket are expected values.

The chi-square statistic is 4.375. The p-value is .112197. The result is not significant at p < .05

The chi-square statistic is 4.375. The p-value is .112197. The result is not significant at p < .01.

We cannot reject the null hypothesis at 5% and 1% Level of significance.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
You have been hired to test whether the demand for a product that your client produces...
You have been hired to test whether the demand for a product that your client produces varies between two demographic markets – the Urban and Rural markets. As such, in each market, you run a short survey that gauges customers demand for your product and assigns them to one of three categories – (i) High (ii) Medium or (iii) Low. You survey 70 people in the “Urban” market and find that their demand falls into the following “buckets” High                32...
You are interested in voting pattern differences between rural and urban voters in the swing state...
You are interested in voting pattern differences between rural and urban voters in the swing state of Iowa. With no big city, you decide that you will categorize counties with more than 50,000 residents as urban and counties with less than 10,000 people as rural. You identify 24 urban counties where 57% of the vote in the last election went for democrats (standard deviation of 11%). And you identify 28 rural counties where 49% of the vote went for democrats...
"A corporation is trying to decide whether to open a new factory outlet store. The store...
"A corporation is trying to decide whether to open a new factory outlet store. The store will have high, medium, or low demand. Before it decides whether to open the store, the corporation can pay $8,000 immediately for a market survey. The market survey can predict 'high,' 'medium,' or 'low' demand. If the actual demand of the store will be high, there is a 0.51 probability the survey will predict 'high' demand, a 0.09 probability the survey will predict 'medium'...
Let x = age in years of a rural Quebec woman at the time of her...
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec gave a sample variance s2 = 3.2. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence...
Let x = age in years of a rural Quebec woman at the time of her...
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec gave a sample variance s2 = 3.1. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence...
A two-way analysis of variance experiment with no interaction is conducted. Factor A has three levels...
A two-way analysis of variance experiment with no interaction is conducted. Factor A has three levels (columns) and Factor B has seven levels (rows). The results include the following sum of squares terms: SST = 346.9 SSA = 196.3 SSE = 79.0 a. Construct an ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "SS" to 2 decimal places, "MS" to 4 decimal places, "F" to 3 decimal places.) Source SS df MS F p-value Rows Columns...
Let x = age in years of a rural Quebec woman at the time of her...
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec gave a sample variance s2 = 3.1. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence...
Let x = age in years of a rural Quebec woman at the time of her...
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 41 women in rural Quebec gave a sample variance s2 = 2.5. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence...
Let x = age in years of a rural Quebec woman at the time of her...
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 41 women in rural Quebec gave a sample variance s2 = 2.4. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence...
Let x = age in years of a rural Quebec woman at the time of her...
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 31 women in rural Quebec gave a sample variance s2 = 2.9. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT