we need more information than that to reprodruce the 2.326. Post the entire question and the working (if known) to get the answer.
The farmer also grows onions. The weight in kilograms of the onions is Normally distributed
with mean 0.155 and variance 0.005. He is trying out a new variety, which he hopes will yield
a higher mean weight. In order to test this, he takes a random sample of 25 onions of the new
variety and ﬁnds that their total weight is 4.77 kg. You should assume that the weight in kilograms
of the new variety is Normally distributed with variance 0.005.
Carry out the test at the 1% level.
Mean weight = 4.77/25 = 0.1908
test statistic= (0.1908-0.155)/sqrt(0.005)*sqrt(25)=2.531
1% level 1-tailed critical value of z = 2.326
2.531 > 2.236 so significant.
There is sufficient evidence to reject H0
sorry, i misread your first post. The 2.326 is the upper 99% point on the normal distribution. You are doing a 1-tailed 1% test so you should look at the 99% point.The farmer also grows onions. The weight in kilograms of the onions is Normally distributed
with mean 0.155 and variance 0.005. He is trying out a new variety, which he hopes will yield
a higher mean weight. In order to test this, he takes a random sample of 25 onions of the new
variety and ﬁnds that their total weight is 4.77 kg. You should assume that the weight in kilograms
of the new variety is Normally distributed with variance 0.005.
Carry out the test at the 1% level.
Mean weight = 4.77/25 = 0.1908
test statistic= (0.1908-0.155)/sqrt(0.005)*sqrt(25)=2.531
1% level 1-tailed critical value of z = 2.326
2.531 > 2.236 so significant.
There is sufficient evidence to reject H0