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t-test, p value and df
發問:
I am analysing my research result by using t-test. I have some questions regarding to the t-test, p value and degree of freedom. >The research is to discover whether the time spent on internet has had any relationship between Group A and Group B. The following is the details of my calculated result:> ... 顯示更多 I am analysing my research result by using t-test. I have some questions regarding to the t-test, p value and degree of freedom. > The research is to discover whether the time spent on internet has had any relationship between Group A and Group B. The following is the details of my calculated result: > Result of Group A: Mean of the minutes spend on internet per day = 58.025 Standard Deviation = 54.4067 Number of people = 40 > Result of Group B: Mean of the minutes spend on internet per day = 46 Standard Deviation = 32.6428 Number of people = 10 > H0: The difference between the mean of minutes spent on Internet per day for Group A and Group B is zero H1: The difference between the mean of minutes spent on Internet per day for Group A and Group B is not zero 1). Can I use t-test to analyse the result if the number of people in two group is different? 2). Is that correct to set the above hypothesis? 3). What is the degree of freedom in the above case? 4). How to find out the p-value by using the t table? > Thank you very much!
1. Yes. You can use t-test to analyse the result 2. Yes. The above hypothesis is correct. 3. You can assume the population variances are the same, and thus, find the pooled variance s. The degree of freedom will be 40+10-2 = 48. 4. You can find the range of p-value without using a computer OR use Z to approximate the p-value. The followings are the steps: s = √{[(40-1)*(54.4067)^2+(10-1)*(32.6428)^2]/(40+10-2)} = √{[(40-1)*(54.4067)^2+(10-1)*(32.6428)^2]/(40+10-2)} = √(125033.4427/48) = √(2604.8634) = 51.0379 t = [(XA_bar-XB_bar)-(μA-μB)]/[s*√(1/nA+1/nB)], where s is the pooled variance = (58.025-46)/[51.0379*√(1/40+1/10)] = 0.6664 = 0.67 df = 48 p-value = 2*P(t > 0.67) p-value is in between 0.4 and 0.6, so we are most likely not to reject H0. In other words, two means are different.
其他解答:
t-test, p value and df
發問:
I am analysing my research result by using t-test. I have some questions regarding to the t-test, p value and degree of freedom. >The research is to discover whether the time spent on internet has had any relationship between Group A and Group B. The following is the details of my calculated result:> ... 顯示更多 I am analysing my research result by using t-test. I have some questions regarding to the t-test, p value and degree of freedom. > The research is to discover whether the time spent on internet has had any relationship between Group A and Group B. The following is the details of my calculated result: > Result of Group A: Mean of the minutes spend on internet per day = 58.025 Standard Deviation = 54.4067 Number of people = 40 > Result of Group B: Mean of the minutes spend on internet per day = 46 Standard Deviation = 32.6428 Number of people = 10 > H0: The difference between the mean of minutes spent on Internet per day for Group A and Group B is zero H1: The difference between the mean of minutes spent on Internet per day for Group A and Group B is not zero 1). Can I use t-test to analyse the result if the number of people in two group is different? 2). Is that correct to set the above hypothesis? 3). What is the degree of freedom in the above case? 4). How to find out the p-value by using the t table? > Thank you very much!
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最佳解答:1. Yes. You can use t-test to analyse the result 2. Yes. The above hypothesis is correct. 3. You can assume the population variances are the same, and thus, find the pooled variance s. The degree of freedom will be 40+10-2 = 48. 4. You can find the range of p-value without using a computer OR use Z to approximate the p-value. The followings are the steps: s = √{[(40-1)*(54.4067)^2+(10-1)*(32.6428)^2]/(40+10-2)} = √{[(40-1)*(54.4067)^2+(10-1)*(32.6428)^2]/(40+10-2)} = √(125033.4427/48) = √(2604.8634) = 51.0379 t = [(XA_bar-XB_bar)-(μA-μB)]/[s*√(1/nA+1/nB)], where s is the pooled variance = (58.025-46)/[51.0379*√(1/40+1/10)] = 0.6664 = 0.67 df = 48 p-value = 2*P(t > 0.67) p-value is in between 0.4 and 0.6, so we are most likely not to reject H0. In other words, two means are different.
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