What is the answer to 809 x 67
2 answers:
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Answer:
The critical values for α = 0.10 are t=±1.7341.
Step-by-step explanation:
The critical values for a two tailed test are the values of t that are placed in the limit of the significance level. If the test statistic t is greater than this critical values, the null hypothesis is rejected.
To determine this values, we first divide the significance level by 2. With a t-table or app, we can determine the t values for P(x>|t|)=α/2=0.05.
In this case, for a degrees of freedom DF=20-2=18, we have the test statistic t=±1.7341 that satisfies the condition.
Solution of the hypothesis test
In this problem we want to prove if the slope of the regression model is different from 0, by means of a hypothesis test.
First, we state the null and alternative hypothesis:
The significance level is .
The degrees of freedom for a simple regression model is:
The standard error estimated of the slope is 0.0053.
The test statistic t is calculated as:
The P-value for a two-tailed test for a t=4.0377 and DF=18 is:
The P-value is smaller than the significance level, so the effect is statistically significant. The null hypothesis is rejected.
Hi!
To compare this two sets of data, you need to use a t-student test:
You have the following data:
-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph
-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph
You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.
To calculate the degrees of freedom you need to use the following equation:
≈34
The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10
So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.
To compare this two sets of data, you need to use a t-student test:
You have the following data:
-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph
-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph
You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.
To calculate the degrees of freedom you need to use the following equation:
The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10
So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.