Question

A local meteorologist claims that his weather forecasts are correct more than 90% of the time. You randomly sample 63 of his past forecasts and find that 52 of them were correct. Perform a test to determine if the meteorologist’s claim is correct. Test at a 10% level of significance.

Please show the formula and the math so I can better understand, thanks

Answer 1

Solution :

Given that,

n = 63

x = 52

= x / n = 52 / 63 = 0.8254

P₀ = 90% = 0.90

1 - P₀ = 1 - 0.90 = 0.10

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.8254 - 0.90 / [(0.90 * 0.10) / 63]

= -1.974

This is the right tailed test .

P(z > -1.974) = 1 - P(z < -1.974) = 1 - 0.0242 = 0.9758

P-value = 0.9758

= 10% = 0.10

P-value >

Fail to reject the null hypothesis .

Meteorologist’s claim is not correct .