A well-known brokerage firm executive claimed that 70% of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that in a sample of 700 people, 74% of them said they are confident of meeting their goals. Test the claim that the proportion of people who are confident is larger than 70% at the 0.05 significance level.

P-value = .010444,  this means it is statistically significant or probable that the proportion of people who are confident is larger than 70% at the 0.05 significance level.

Explanation:

Population proportion, p = 0.70

Number of people taken in a sample, n = 700

Sample proportion, = 0.74

State the hypotheses:

H ₀:p = 0.70

Hₐ :p < 0.70

one-tailed test, we must calculate z test statistic:

z = ( - p) / √{[p(1 - p)]/n}

z = (0.74 - 0.70) / √{[0.70(1 - 0.70)]/700} = 2.31

z = 2.31

using a table we can find out P-value for Z = 2.31:

P-value = .010444,  this means it is statistically significant or probable that the proportion of people who are confident is larger than 70% at the 0.05 significance level.

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the answer is: c.that person might be arrested and deported

filling w-4 form incorrectly usually would rise suspicions from government's auditors. when they notice such mistakes, they would send their teams to investigate the situation further.

if those investigation teams believe that the incorrectness is done on purpose (with malicious intent), that person can be arrested or deported