question archive Can you please help I am stuck on questions like this one
Can you please help I am stuck on questions like this one
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Can you please help I am stuck on questions like this one. In a sample of 75 planters mixed nuts, 12 were almonds. Calculate a 90% confidence interval for the true proportion of almonds.
The 90% confidence interval is from to
What sample size would be needed for 90% confidence and an error of + .03
sample size
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Given;
n=75
x= 12
The sample proportion is P=x/n
12/75= 0/16
a)
The 90% confidence interval
P(+-)z√P(1-p)/n
0.16(+-)(1.645)√0.16(1-0.16)/75
using the z table = 1.645
0.16(+-)1.645√0.16(0.84)/75
0.16(+-)(1.645)√0.1344/75
So;
=0.16(+-)(1.645)√1.792*10^-3
=0.16(+-)(1.645)(0.042332)
=0.16(+-) 0.06963614
=(0.0904, 0.2296)
The 90% confidence interval is from;
0.0904 to 0.2296.
b)
Given= 0.03
z=1.645
Now, sample size n= (n/E)2p(1-p)
=(1.645/0.03)2(0.16)(1-0.16)
=(54.833)2(0.16)(0.84)
=3006.6907*0.1344
=404.099
n= 4040
Step-by-step explanation
Given;
n=75
x= 12
The sample proportion is P=x/n
12/75= 0/16
a)
The 90% confidence interval
P(+-)z√P(1-p)/n
0.16(+-)(1.645)√0.16(1-0.16)/75
using the z table = 1.645
0.16(+-)1.645√0.16(0.84)/75
0.16(+-)(1.645)√0.1344/75
So;
=0.16(+-)(1.645)√1.792*10^-3
=0.16(+-)(1.645)(0.042332)
=0.16(+-) 0.06963614
=(0.0904, 0.2296)
The 90% confidence interval is from;
0.0904 to 0.2296.
b)
Given= 0.03
z=1.645
Now, sample size n= (n/E)2p(1-p)
=(1.645/0.03)2(0.16)(1-0.16)
=(54.833)2(0.16)(0.84)
=3006.6907*0.1344
=404.099
n= 404
The sample size is 404.
