A national poll reported that 58% of those with Internet access have made purchases online. To investigate whether this percentage applies to its own state, a legislator commissions a study. A random sample of 400 state residents who have Internet access is taken. Of those 400 respondents, 215 said that they have made purchases online. Does this sample provide sufficient evidence to conclude that the state differs from the nation with respect to making purchases online? Suppose the significance level is 0.05.

Null Hypothesis is accepted

Conclusion : proportion of people with internet access having made online purchase = 58%

Step-by-step explanation:

Let proportion of people with internet access having made online purchase = x

Null Hypothesis [H0] : x = 58% = 0.58

Alternate Hypothesis [H1] : x ≠ 58% ≠ 0.58

Observed x proportion (mean) : x' = 215 / 400 = 0.5375

z statistic = [ x' - x ]  / √ [ { x ( 1-x ) } / n ]

( 0.5375 - 0.58 ) /  √ [ 0.58 (0.42) / 400 ]

= -0.0425 / √ 0.000609

= -0.0425  / 0.02467

z = - 1.72

Since calculated value of z, 1.72 < tabulated value of z at 0.05 significance level, 1.96

So, we accept the null hypothesis & conclude that x = 58%

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6

the answer will be