91Ó°ÊÓ

Given in the question that, the each batch of mixed nuts contains $52\%$cashews, $27\%$almonds, $13\%$macadamia nuts, and $8\%$brazil nuts.

$H_0$: will be the null hypothesis. The observed count of each type of nut in a group of $150$nuts from the most recent batch matches the reported count of nuts in a group of $150$

$H_0:p_c=52\%,p_a=27\%,p_m=13\%,p_b=8\%$

$H_1$: The observed counts of each type of nut in a group of $150$nuts from the most recent batch do not match the reported count of nuts in a $150$group.

Because the claim specifies a proportion of each type of nut of $100$, we can calculate the proportion of each type of nut in 150 by multiplying each number by $150.$

Given in the question that,

According to the information,

Here,

$n=150$

$P\left(\text{cashews}\right)=52\%=0.52$

$P\left(\text{almonds}\right)=27\%=0.27$

$P\left(\text{Macadamia}\right)=13\%=0.13$

$P\left(brazil\right)=8\%=0.08$

The expected count is then calculated as the product of the sample size $n$and the type probability:

Cashews:

$150(0.52)=78$

Almonds:

$150(0.27)=40.5$

Macadamia:

$150(0.13)=19.5$

Brazil:

$150(0.08)=12$

Therefore, the expected counts of nuts are

$\begin{array}{|cccc|}\hline \text{Cashews}& \text{Almonds}& \text{Macadamia}& \text{Brazil}\\ 78& 40.5& 19.5& 12\\ \hline\end{array}$