Decide whether the normal sampling distribution can be used. If it can be used, test the claim about the population proportion p at the given level of significance alpha using the given sample statistics. Claim: p is not equal to 0.22; Alpha = 0.01; Sample statistics: p = 0.15, n = 180Can the normal sampling distribution be used? A. No, because np is less than 5. B. No, because nq is less than 5. C. Yes, because both np and nq are greater than or equal to 5. D. Yes, because pq is greater than alpha = 0.01. State the null and alternative hypotheses. Determine the critical value(s).Find the z-test statistic.

A

   The correct option is  C

B

  The  value of  z-test is  

Step-by-step explanation:

From the question we are told that  

    The null hypothesis  

    The alternative hypothesis  

    The  level of significance is  

    The sample proportion is  

     The sample size is  n = 180

Generally from central limit theorem

if   np  and  nq  are  >  5  then  normal sampling distribution can be used

So  

     np =  180  * 0.22 =  39.6  >  5

and  

     nq =  180 * (1 -0.22) =  140.4 > 5

So normal sampling distribution can be used

Generally the z-test is mathematically represented as

=>  

=>  

answer: if you weigh 60 pounds on earth, on the moon you will weigh about 10 pounds. if you weigh 120 pounds on earth, you will weigh about 20 pounds on the moon. since weight is caused by the mass of two objects that are near each other, the bigger the objects, the more the force of gravity.

step-by-step explanation: if you weigh 60 pounds on earth, on the moon you will weigh about 10 pounds. if you weigh 120 pounds on earth, you will weigh about 20 pounds on the moon. since weight is caused by the mass of two objects that are near each other, the bigger the objects, the more the force of gravity.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47