Question

Pyramid Lake is on the Paiute Indian Reservation in Nevada. The lake is famous for cutthroat trout.      Suppose a friend tells you that the mean length of trout caught in Pyramid Lake is 19 inches. However, the Creel Survey reported that of a random sample of 51 fish caught, the mean length was 18.5 inches, with      standard deviation of 3.2 inches. Using a 10% level of significance, should you ‘accept’ or reject your      friend’s claim?

Answer 1

Given,

$sample \; mean,\bar{x} = 18.5$

$sample \; standard\;deviation,\s = 3.2$

The hypothesis is,

$H_0 : \bar{x} = 19$

$H_0 : \bar{x} \neq 19$

we know that the z test statistic is,

$z= \frac{\bar{x}- 19 }{ \frac{\sigma} {\sqrt{n}}}$

$z= \frac{18.5- 19 }{ \frac{3.2} {\sqrt{51}}}$

$z=-1.1158$

The corresponding p value is,

p = 0.2645

Conclusion : Since the p value is greater than the level of significance(0.1), we fail to reject the null hypothesis and accept the friends claim