Question

Claim: Most adults would not erase all them would erase all of their personal information online if they could. Make a subjective estimate to decide whether the results are significantly low or significantly high then state a conclusion about the original claim their personal information online if they could. A software firm survey of 615 randomly selected adults showed that 0.6% of significantly so there The results sufficient evidence to support the claim that most adults would not erase all of their personal information online if they could 2 of 2 of o omplet of oe omplete Click select your answer(s) Us hp HEWLE

Answer 1

GIVEN:

Sample size

(n) = 615

Sample proportion of randomly selected adults who erase all of their personal information online

(Phat = 0.60

HYPOTHESIS:

(That is, the proportion of adults who erase all of their personal information online is 50% (or) Most adults would not erase all of their personal information online.)

Ho p 0.50

(That is, the proportion of adults who erase all of their personal information online is more than 50% (or) Most adults would erase all of their personal information online.)

HI p>0.50

LEVEL OF SIGNIFICANCE:
0.05

TEST STATISTIC:

(Phat Po)/po(1 - po))/n1/2

which follows standard normal distribution.

where

Po
is the hypothesized value
0.50
.

CALCULATION:

(Phat Po)/po(1 - po))/n1/2

  $=(0.60-0.50)/[(0.50(1-0.50))/615]^{1/2}$

  
= 0.1/0.02016

Z= 4,96

P VALUE:

The p value for given right tailed test (since ) is,

HI p>0.50

$P[z>4.96]=1-P[z<4.96]$

From the z table, the probability value is the value with corresponding row 4.9 and column 0.06.

  

=1-0.9999

P 4.96 0.0001

DECISION RULE:

$Reject\, \, H_{0}\, \, if\, \, p-value\leq \alpha$

CONCLUSION:

Since the calculated p value (0.0001) is less than the significance level , we reject null hypothesis and conclude that the proportion of adults who erase all of their personal information online is more than 50%. In other words, most adults would erase all of their personal information online.

0.05

Thus the results are significantly true so there is no sufficient evidence to support the claim that most adults would not erase all of their personal information online is they could.

Answer 2

GIVEN:

Sample size

(n) = 615

Sample proportion of randomly selected adults who erase all of their personal information online

(Phat = 0.60

HYPOTHESIS:

(That is, the proportion of adults who erase all of their personal information online is 50% (or) Most adults would not erase all of their personal information online.)

Ho p 0.50

(That is, the proportion of adults who erase all of their personal information online is more than 50% (or) Most adults would erase all of their personal information online.)

HI p>0.50

LEVEL OF SIGNIFICANCE:
0.05

TEST STATISTIC:

(Phat Po)/po(1 - po))/n1/2

which follows standard normal distribution.

where

Po
is the hypothesized value
0.50
.

CALCULATION:

(Phat Po)/po(1 - po))/n1/2

  $=(0.60-0.50)/[(0.50(1-0.50))/615]^{1/2}$

  
= 0.1/0.02016

Z= 4,96

P VALUE:

The p value for given right tailed test (since ) is,

HI p>0.50

$P[z>4.96]=1-P[z<4.96]$

From the z table, the probability value is the value with corresponding row 4.9 and column 0.06.

  

=1-0.9999

P 4.96 0.0001

DECISION RULE:

$Reject\, \, H_{0}\, \, if\, \, p-value\leq \alpha$

CONCLUSION:

Since the calculated p value (0.0001) is less than the significance level , we reject null hypothesis and conclude that the proportion of adults who erase all of their personal information online is more than 50%. In other words, most adults would erase all of their personal information online.

0.05

Thus the results are significantly true so there is no sufficient evidence to support the claim that most adults would not erase all of their personal information online is they could.