BigDeal Real Estate surveyed prices per square foot in the valley and foothills of Hoke-a mo, Utah. Based on BD’s DATA, are process per square foot equal at x=0.01?
Valley |
Foothills |
109 |
82 |
116 |
161 |
106 |
163 |
157 |
112 |
147 |
222 |
105 |
137 |
173 |
226 |
153 |
200 |
137 |
154 |
110 |
176 |
The critical value is 2.977 since this is a two-tail scenario. The test statistic is 2.239. Since the test statistic < the critical value, the test statistic does not lie in the area of rejection. Do not reject the null hypothesis. The prices per square foot are equal at alpha= .01
The critical value is 2.977 since this is a two-tail scenario. The test statistic is 1.513. Since the test statistic < the critical value, the test statistic does not lie in the area of rejection. Do not reject the null hypothesis. The prices per square foot are equal at alpha= .01
The critical value is 2.977 since this is a two-tail scenario. The test statistic is 1.936. Since the test statistic < the critical value, the test statistic does not lie in the area of rejection. Do not reject the null hypothesis. The prices per square foot are equal at alpha= .01
The critical value is 2.977 since this is a two-tail scenario. The test statistic is 3.207. Since the test statistic > the critical value, the test statistic does not lie in the area of rejection. Reject the null hypothesis. The prices per square foot are equal at alpha= .01
Here we have data :
Calley | Foothills |
109 | 82 |
116 | 161 |
106 | 163 |
157 | 112 |
147 | 222 |
105 | 137 |
173 | 226 |
153 | 200 |
137 | 154 |
110 | 176 |
Here we are using Excel for calculation:
t-Test: Two-Sample Assuming Unequal Variances | ||
Variable 1 | Variable 2 | |
Mean | 131.3 | 163.3 |
Variance | 629.5667 | 2101.1222 |
Observations | 10 | 10 |
Hypothesized Mean Difference | 0 | |
df | 14 | |
t Stat | -1.9365 | |
P(T<=t) two-tail | 0.0733 | |
t Critical two-tail | 2.977 |
Correct option:
The critical value is 2.977 since this is a two-tail scenario. The test statistic is 1.936. Since the test statistic < the critical value, the test statistic does not lie in the area of rejection. Do not reject the null hypothesis. The prices per square foot are equal at alpha= .01
BigDeal Real Estate surveyed prices per square foot in the valley and foothills of Hoke-a mo,...
BigDeal Real Estate surveyed prices per square foot in the valley and foothills of Hoke-a-mo, Utah. Based on BD's DATA, are prices per square foot equal at beta= 0.01? 1. The critical value is 2.977 since this is a two-tail scenario. The test statustuc us 3.207. Since the test statistic> the critical value, the test statistic does lie in the area of rejection. Reject the null hypothesis. The prices per square foot are not equal at alpha = .01 2....
A real estate agent has claimed that the average home prices in Lindville (L) and Jamesville (J) are equal. Suppose that a random sample of 18 Lindville homes resulted in a sample mean price of $235,000 with a sample standard deviation of $22,000. A random sample of 12 Jamesville homes resulted in a sample mean price of $245,000 with a sample standard deviation of $24,000. Test the real estate agent's claim at the 5% significance level. Assume that population variances...
According to Relators-R-Us, the average price per square foot of homes in western Kansas is $85 with a standard deviation of $11.24. Using recent sales figures that happened in the city of Hays, the relators wish to determine if the average price per square foot in Hays is significantly different at the 1% level. In using a collection of thirty-seven randomly homes recently sold in Hays, they found the average price of these home was at $88.25 per square foot....
Test using the p-value approach with ? = 0.05.State the null and alternative hypothesis.H0: ? < 98.6 versus Ha: ? > 98.6H0: ? = 98.6 versus Ha: ? > 98.6 H0: ? = 98.6 versus Ha: ? < 98.6H0: ? = 98.6 versus Ha: ? ≠ 98.6H0: ? ≠ 98.6 versus Ha: ? = 98.6Find the test statistic and the p-value. (Round your test statistic to two decimal places and your p-value to four decimal places.)z=p-value=State your conclusion.The p-value is greater than alpha so H0 is not rejected. There is insufficient evidence to indicate that the average body temperature for healthy humans deviates from 98.6°.The p-value is less than alpha so H0 is rejected. There is sufficient evidence to...
(a)Test using the p-value approach with ? = 0.05.State the null and alternative hypothesis.H0: ? < 98.6 versus Ha: ? > 98.6H0: ? = 98.6 versus Ha: ? > 98.6 H0: ? = 98.6 versus Ha: ? < 98.6H0: ? = 98.6 versus Ha: ? ≠ 98.6H0: ? ≠ 98.6 versus Ha: ? = 98.6Find the test statistic and the p-value. (Round your test statistic to two decimal places and your p-value to four decimal places.)z=p-value=State your conclusion.The p-value is greater than alpha so H0 is not rejected. There is insufficient evidence to indicate that the average body temperature for healthy humans deviates from 98.6°.The p-value is less than alpha so H0 is rejected. There is sufficient evidence to...
a) true b) false 42. For a chi-square distributed random variable with 10 degrees of freedom and a level of sigpificanoe computed value of the test statistics is 16.857. This will lead us to reject the null hypothesis. a) true b) false 43. A chi-square goodness-of-fit test is always conducted as: a. a lower-tail test b. an upper-tail test d. either a lower tail or upper tail test e. a two-tail test 44. A left-tailed area in the chi-square distribution...
Consider the following hypothesis statement using alpha equals0.05 and data from two independent samples. Assume the population variances are equal and the populations are normally distributed. Complete parts a and b. Upper H 0 : mu 1 minus mu 2 equals 0 x overbar 1 equals 14.7 x overbar 2 equals 12.0 Upper H 1 : mu 1 minus mu 2 not equals 0 s 1 equals 2.7 s 2 equals 3.3 n 1 equals 20 n 2 equals 15...
READ THE DIRECTIONS PLEASE 6) Make your own word problem of the form of this problem. Your word problem should apply directly to biology major. State the null and alternative hypotheses and alpha (α) . (Since you don’t have any data, you need not actually conduct the hypothesis test). Please make a word problem stated above and answer all the parts based on the question below. 2 A data set includes data from student evaluations of courses. The summary statistics...
10. The owner of a company has asked you to conduct an evaluation of the customer satisfaction ratings to see if the company continues to provide customers with customer service that ranks above average. To be considered above average the average customer satisfaction score has to be above 7. Suppose a random sample of 60 customers is taken from a population to evaluate customer satisfaction. The sample mean is 7.25. The sample standard deviation is 1.05. The population mean is...
Hypothesis Testing Method A = ADKAR Framework (column A) Method B = Prosci Change Management Methodology (column B) Null hypothesis is H0: Method A = Method B Research (Alternative) hypothesis is H1:Method A < Method B Sample size: 30 One-tailed test: Direction stated in the hypothesis is determine more effective (greater or less than). Level of Significance (a): .05 t-test is used because standard deviation of population is unknown, sample size is less than 30 Cutoff Sample Score (critical value)...