Under Ho: Bweight = =0 against Ha: Bweight #0 to test the significance of the variable Weight; what are the values of the test statistic and the P-value of the test...

Find the P-value for the hypothesis test with the standardized test statistic $z .$ Decide whether to reject $H_{0}$ for the level of significance $\alpha .$ Right-tailed test $z=1.23$ $\alpha=0.10$

Now. This time this is a two tailed test. My Z value is minus 1.68 and my AL 50.5 So minus 1.68 this is my Z value, my Z statistic, and my Alfa is 0.5 Now, when I'm finding the p value, I will put in reserve statistic and I will choose this to be two tailed and my P value comes out to be 0.930 My P value is turning out to be 0.930 And since my P value is greater than my Alfa, I will fail to reject final hypothesis and this is going to be my answer.

Find the P value for the hypothesis test with the standardized test statistic Z decide whether the reject edge not for the significant Several Alfa. Now, this is a left tail test where my Z statistic is minus 1.32 and my AL 50.1. So this is minus 1.32 This is my very statistic. And my Alfa is equal to 0.1. So what do I do? I simply go to a P value calculator. I put in my Zed school, which is minus one point, which is minus 1.32 So this will be minus 1.32 and this is a left field s. So I'll just right here one tail and my P value in this case is zero point 0934 I find that I my p values 0.93 For instance, this is less than 0.1. My people live is 0.934 And since this is less than Alfa, I reject mine al hypothesis. I will reject my null hypothesis and this will be my answer

We're conducting a left tailed hypothesis test for population me and you. Mhm. And we have a sample with n equals 10. And after using the t. Distribution to compute the test statistic, obtained test statistic t equals negative 3.381. We want to answer the following questions based on this setup. First, we want to use a tea table to find the key value in a tea table. We will find the T. Interval for this T stat For degree of freedom and -1 equals nine. Is that two years less than negative? 3.250 from the same tea table. We found that this maps directly onto a p value Of P less than .005. We can now use that information to answer part B. At what significance levels can or can't we reject the null hypothesis H. Nine. To answer this, we remember that we can reject H. Not whenever P. Is less than people to alpha. So we can definitely reject h. Not when alpha is greater than or equal 2.005. But we failed to reject for P less than 0.5 or rather alpha less than 0.5 because we do not know the precise P value as the tea table does not provide a precise measurement of P. For this T value.

Yeah. When conducting a right tail hypothesis test for a population meaning you will sample size and equals 20 and output a T statistic 2.235 That is we're conducting a T. Test. We want to answer the following questions first. We want to use a tea table to find the P value for this test. In order to do that we have to find the T interval that artie value falls between for a degree of freedom 19. This is 2.93 is less than T. Is less than 2.539 From the tea table. This map directly onto AP. Value of .01 is less than P is less than .025. Now we enter based on this at what significance levels can we or can we reject the null hypothesis? H mart To answer this, we have to remember that we can reject the null hypothesis. It's not for wherever P is less than or equal to alpha. So based off RP interval, we can reject for any alpha greater than or equal 2.025. But we failed to reject whenever alpha is less than .01. We can't specifically say about anything in between these p values unless we use a more formal way of calculating our P value.