Question
how can I decide to reject or fail to reject the following null hypothesis? let a= 0.05,data; p=0.69, n=32, test statistic z= 0.09095
H(0);p=0.68 H(1)p>0.68
H(0);p=0.68 H(1)p>0.68
Answers
To decide whether to reject or fail to reject the null hypothesis, we would compare the test statistic (z) to the critical value corresponding to the significance level (α).
First, determine the critical value using the significance level (α) = 0.05. Since this is a one-tailed test (H1: p > 0.68), we need to find the z-value related to the 0.05 in the tail. You can look up this value in a standard normal distribution table or use a calculator with a built-in function. The z-value corresponding to α = 0.05 in a one-tailed test is 1.645.
Now compare the test statistic (z) to the critical value:
Test statistic, z = 0.09095
Critical value = 1.645
Since the test statistic (0.09095) is less than the critical value (1.645), we fail to reject the null hypothesis.
In conclusion, there is not enough evidence to support the alternative hypothesis (p > 0.68) at a significance level of 0.05.
First, determine the critical value using the significance level (α) = 0.05. Since this is a one-tailed test (H1: p > 0.68), we need to find the z-value related to the 0.05 in the tail. You can look up this value in a standard normal distribution table or use a calculator with a built-in function. The z-value corresponding to α = 0.05 in a one-tailed test is 1.645.
Now compare the test statistic (z) to the critical value:
Test statistic, z = 0.09095
Critical value = 1.645
Since the test statistic (0.09095) is less than the critical value (1.645), we fail to reject the null hypothesis.
In conclusion, there is not enough evidence to support the alternative hypothesis (p > 0.68) at a significance level of 0.05.
Related Questions
In a chi-squared test of a contingency table, the value of the test statistic was =12.678, and the...
Use the given information to find the P-value. Also, use a 0.05 significance level and state the con...
Based on a sample of 40 people, the sample mean GPA was 3.18 with a standard deviation of 0.06
Th...
Use the given information to find the P-value. Also, use a 0.05 significance level and state the con...