As per the given standard deviation, the p - value is 0.09997
The term standard deviation means a number that represents the "spread" or "dispersion" of a set of data and there are other measures for spread, such as range and variance.
Here we have given that teacher believes that the standard deviation of scores for a particular test that he gives every semester is 4 points. His current students claim that the standard deviation is more than 4 points. Let σ2 be the true variance for the test scores.
Here we need to find the P-value for the appropriate test, rounded to 3 decimal places.
While we looking into the given question we have identified the values of the following
Standard deviation = 5.2
H0 : σ² = 16
H1 : σ² > 16.
Now, we have to calculate the value of z as,
Let us consider that the significance level as 0.05, then the critical value of z is,
Critical value (1-tailed) = 1.645.
Then the p - value is calculated as,
Based on the critical value, the P-Value is .09997.
And the result is not significant at p < .05.
To know more about Standard deviation here.
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